Download Determine whether each sequence is arithmetic no. 1. 8, –2, –12

 10-1 Sequences as Functions
ANSWER: Yes
Determine whether each sequence is arithmetic
no. 1. 8, –2, –12, –22
3. 1, 2, 4, 8, 16
SOLUTION: Subtract each term from the term directly after it.
SOLUTION: Subtract each term from the term directly after it.
There is no common difference.
Therefore, the sequence is not arithmetic.
The common difference is –10.
Therefore, the sequence is arithmetic.
ANSWER: No
ANSWER: Yes
4. 0.6, 0.9, 1.2, 1.8, ...
2. –19, –12, –5, 2, 9
SOLUTION: Subtract each term from the term directly after it.
SOLUTION: Subtract each term from the term directly after it.
The common difference is 7.
Therefore, the sequence is arithmetic.
ANSWER: Yes
3. 1, 2, 4, 8, 16
SOLUTION: Subtract each term from the term directly after it.
There is no common difference.
Therefore, the sequence is not arithmetic.
ANSWER: No Manual - Powered by Cognero
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4. 0.6, 0.9, 1.2, 1.8, ...
There is no common difference.
Therefore, the sequence is not arithmetic.
ANSWER: No
Find the next four terms of each arithmetic
sequence. Then graph the sequence.
5. 6, 18, 30, …
SOLUTION: Subtract each term from the term directly after it.
The common difference is 12.
Therefore, the sequence is arithmetic.
To find the next term, add 12 to the last term.
30 + 12 = 42
42 + 12 = 54
54 + 12 = 66
66 + 12 = 78
Page 1
ANSWER: 10-1No
Sequences as Functions
Find the next four terms of each arithmetic
sequence. Then graph the sequence.
5. 6, 18, 30, …
6. 15, 6, –3, …
SOLUTION: Subtract each term from the term directly after it.
SOLUTION: Subtract each term from the term directly after it.
The common difference is –9.
Therefore, the sequence is arithmetic.
The common difference is 12.
Therefore, the sequence is arithmetic.
To find the next term, add –9 to the last term.
–3 + (–9) = –12
–12 + (–9) = –21
–21 + (–9) = –30
–30 + (–9) = –39
To find the next term, add 12 to the last term.
30 + 12 = 42
42 + 12 = 54
54 + 12 = 66
66 + 12 = 78
Graph the sequence.
Graph the sequence.
ANSWER: –12, –21, –30, –39
ANSWER: 42, 54, 66, 78
7. –19, –11, –3, …
6. 15, 6, –3, …
SOLUTION: Subtract each term from the term directly after it.
SOLUTION: eSolutions
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Subtract each term from the term directly after it.
The common difference is 8.
Therefore, the sequence is arithmetic.
Page 2
10-1 Sequences as Functions
7. –19, –11, –3, …
8. –26, –33, –40, …
SOLUTION: Subtract each term from the term directly after it.
SOLUTION: Subtract each term from the term directly after it.
The common difference is 8.
Therefore, the sequence is arithmetic.
The common difference is –7.
Therefore, the sequence is arithmetic.
To find the next term, add 8 to the last term.
To find the next term, add –7 to the last term.
–3 + 8 = 5
5 + 8 = 13
13 + 8 = 21
21 + 8 = 29
–40 + (–7) = –47
–47 + (–7) = –54
–54 + (–7) = –61
–61 + (–7) = –68
Graph the sequence.
Graph the sequence.
ANSWER: 5, 13, 21, 29
ANSWER: –47, –54, –61, –68
8. –26, –33, –40, …
SOLUTION: Subtract each term from the term directly after it.
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The common difference is –7.
9. FINANCIAL LITERACY Kelly is saving her
money to buy a car. She has $250, and she plans to
save $75 per week from her job as a waitress.
a. How much will Kelly have saved after 8 weeks?
b. If the car costs $2000, how long will it take her to
Page 3
save enough money at this rate?
a. $850
10-1 Sequences as Functions
9. FINANCIAL LITERACY Kelly is saving her
money to buy a car. She has $250, and she plans to
save $75 per week from her job as a waitress.
a. How much will Kelly have saved after 8 weeks?
b. If the car costs $2000, how long will it take her to
save enough money at this rate?
SOLUTION: a. Given a 0 = 250, d = 75 and n = 8.
After 8 weeks, she will have 250 + (8 × 75) or $850.
b. Given a n = 2000.
Find n.
b. 24 wk
Determine whether each sequence is
geometric. Write yes or no.
10. –8, –5, –1, 4, …
SOLUTION: Find the ratio of the consecutive terms.
Since the ratios are not same, the sequence is not
geometric.
ANSWER: No
11. 4, 12, 36, 108, …
So, it will take about 24 weeks to save $2000.
ANSWER: a. $850
b. 24 wk
Determine whether each sequence is
geometric. Write yes or no.
10. –8, –5, –1, 4, …
SOLUTION: Find the ratio of the consecutive terms.
SOLUTION: Find the ratio of the consecutive terms.
Since the ratios are the same, the sequence is
geometric.
ANSWER: Yes
12. 27, 9, 3, 1, …
SOLUTION: Find the ratio of the consecutive terms.
Since the ratios are not same, the sequence is not
geometric.
ANSWER: No
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Since the ratios are the same, the sequence is
geometric.
ANSWER: Yes
11. 4, 12, 36, 108, …
Page 4
ANSWER: 10-1Yes
Sequences as Functions
Since the ratios are the same, the sequence is
geometric
To find the next term, multiply the previous term by
12. 27, 9, 3, 1, …
.
SOLUTION: Find the ratio of the consecutive terms.
Since the ratios are the same, the sequence is
geometric.
Graph the sequence.
ANSWER: Yes
13. 7, 14, 21, 28, …
SOLUTION: Find the ratio of the consecutive terms.
Since the ratios are not the same, the sequence is not
geometric.
ANSWER: 40.5, 60.75, 91.125
ANSWER: No
Find the next three terms of each geometric
sequence. Then graph the sequence.
14. 8, 12, 18, 27, …
SOLUTION: Find the ratio of the consecutive terms.
15. 8, 16, 32, 64, …
SOLUTION: Find the ratio of the consecutive terms.
Since the ratios are the same, the sequence is
geometric
To find the next term, multiply the previous term by
.
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Since the ratios are the same, the sequence is
geometric.
To find the next term, multiply the previous termPage
by 5
2.
16. 250, 50, 10, 2, …
10-1 Sequences as Functions
15. 8, 16, 32, 64, …
SOLUTION: Find the ratio of the consecutive terms.
SOLUTION: Find the ratio of the consecutive terms.
Since the ratios are the same, the sequence is
geometric
To find the next term, multiply the previous term by
Since the ratios are the same, the sequence is
geometric.
.
To find the next term, multiply the previous term by
2.
Graph the sequence.
Graph the sequence.
ANSWER: 128, 256, 512
ANSWER: 16. 250, 50, 10, 2, …
SOLUTION: Find the ratio of the consecutive terms.
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17. 9, –3, 1,
,…
Page 6
ANSWER: 10-1 Sequences as Functions
17. 9, –3, 1,
,…
SOLUTION: Find the ratio of the consecutive terms.
Determine whether each sequence is
arithmetic, geometric, or neither. Explain your
reasoning.
To find the next term, multiply the previous term by
Since the ratios are the same, the sequence is
geometric.
18. 5, 1, 7, 3, 9, …
.
SOLUTION: There is no common difference.
Therefore, the sequence is not arithmetic.
Graph the sequence.
Find the ratio of the consecutive terms.
Since the ratios are not the same, the sequence is not
geometric.
ANSWER: Neither; there is no common difference or ratio.
19. 200, –100, 50, –25, …
ANSWER: SOLUTION: To find the common difference, subtract any term
from the term directly after it.
There is no common difference.
Therefore, the sequence is not arithmetic.
eSolutions Manual - Powered by Cognero
Find the ratio of the consecutive terms.
Page 7
ANSWER: ANSWER: Neither; there is no common difference or ratio.
10-1 Sequences as Functions
19. 200, –100, 50, –25, …
Geometric; the common ratio is
.
20. 12, 16, 20, 24, …
SOLUTION: To find the common difference, subtract any term
from the term directly after it.
SOLUTION: To find the common difference, subtract any term
from the term directly after it.
There is no common difference.
Therefore, the sequence is not arithmetic.
The common difference is 4.
Therefore, the sequence is arithmetic.
Find the ratio of the consecutive terms.
Find the ratio of the consecutive terms.
The common ratio is
Since the ratios are not the same, the sequence is not
geometric.
.
Since the ratios are the same, the sequence is
geometric.
ANSWER: Arithmetic; the common difference is 4.
ANSWER: Geometric; the common ratio is
Determine whether each sequence is
arithmetic. Write yes or no.
.
20. 12, 16, 20, 24, …
21. SOLUTION: To find the common difference, subtract any term
from the term directly after it.
SOLUTION: Subtract any term from the term directly after it.
The common difference is 4.
Therefore, the sequence is arithmetic.
Find the ratio of the consecutive terms.
There is no common difference.
Therefore, the sequence is not arithmetic.
ANSWER: No
Since the ratios are not the same, the sequence is not
geometric.
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eSolutions
ANSWER: 22. –9, –3, 0, 3, 9
SOLUTION: Page 8
ANSWER: No
10-1 Sequences as Functions
ANSWER: No
22. –9, –3, 0, 3, 9
24. SOLUTION: Subtract any term from the term directly after it.
SOLUTION: Subtract any term from the term directly after it.
There is no common difference.
Therefore, the sequence is not arithmetic.
ANSWER: No
The common difference is
.
Therefore, the sequence is arithmetic.
23. 14, –5, –19, …
ANSWER: Yes
SOLUTION: Subtract any term from the term directly after it.
Find the next four terms of each arithmetic
sequence. Then graph the sequence.
25. –4, –1, 2, 5,…
There is no common difference.
Therefore, the sequence is not arithmetic.
SOLUTION: Subtract any term from the term directly after it.
ANSWER: No
The common difference is 3.
Therefore, the sequence is arithmetic.
24. To find the next term, add 3 to the last term.
SOLUTION: Subtract any term from the term directly after it.
5+3=8
8 + 3 = 11
11 + 3 = 14
14 + 3 = 17
Graph the sequence.
The common difference is
.
Therefore, the sequence is arithmetic.
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ANSWER: Yes
Page 9
ANSWER: 10-1Yes
Sequences as Functions
Find the next four terms of each arithmetic
sequence. Then graph the sequence.
25. –4, –1, 2, 5,…
26. 10, 2, –6, –14, …
SOLUTION: Subtract any term from the term directly after it.
SOLUTION: Subtract any term from the term directly after it.
The common difference is –8.
Therefore, this sequence is arithmetic.
The common difference is 3.
Therefore, the sequence is arithmetic.
To find the next term, add –8 to the last term.
–14 + (–8) = –22
–22 + (–8) = –30
–30 + (–8) = –38
–38 + (–8) = –46
To find the next term, add 3 to the last term.
5+3=8
8 + 3 = 11
11 + 3 = 14
14 + 3 = 17
Graph the sequence.
Graph the sequence.
ANSWER: –22, –30, –38, – 46
ANSWER: 8, 11, 14, 17
27. –5, –11, –17, –23, …
26. 10, 2, –6, –14, …
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eSolutions
SOLUTION: Subtract any term from the term directly after it.
SOLUTION: Subtract any term from the term directly after it.
Page 10
10-1 Sequences as Functions
27. –5, –11, –17, –23, …
28. –19, –2, 15, …
SOLUTION: Subtract any term from the term directly after it.
SOLUTION: Subtract any term from the term directly after it.
The common difference is –6.
Therefore, the sequence is arithmetic.
The common difference is 17.
Therefore, the sequence is arithmetic.
To find the next term, add –6 to the last term.
To find the next term, add 17 to the last term.
–23 + (–6) = –29
–29 + (–6) = –35
–35 + (–6) = –41
–41 + (–6) = –47
15 + 17 = 32
32 + 17 = 49
49 + 17 = 66
66 + 17 = 83
Graph the sequence.
Graph the sequence.
ANSWER: 32, 49, 66, 83
ANSWER: –29, –35, – 41, – 47
29. 28. –19, –2, 15, …
SOLUTION: eSolutions
Manual - Powered by Cognero
Subtract any term from the term directly after it.
SOLUTION: Page 11
Subtract any term from the term directly after it.
ANSWER: 29. 10-1 Sequences as Functions
SOLUTION: Subtract any term from the term directly after it.
The common difference is
Therefore, the sequence is arithmetic.
To find the next term, add
.
to the last term.
30. SOLUTION: Subtract any term from the term directly after it.
The common difference is –1.
Therefore, the sequence is arithmetic.
Graph the sequence.
To find the next term, add −1 to the last term.
ANSWER: Graph the sequence.
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ANSWER: Page 12
b. On what day will Mario first row an hour or
more?
10-1 Sequences as Functions
c. Is it reasonable for this pattern to continue
indefinitely? Explain.
ANSWER: SOLUTION: a. Given a 1 = 5, d = 1.5 and n = 18.
Find a 18.
Therefore he will row for 30 minutes and 30 seconds
on the 38th day.
31. THEATER There are 28 seats in the front row of a
theater. Each successive row contains two more
seats than the previous row. If there are 24 rows,
how many seats are in the last row of the theater?
b. Given a 1 = 5, d = 1.5 and a n = 60.
Find n.
SOLUTION: Given a 1 = 28, d = 2 and n = 24.
Find a 24.
Mario will first row an hour or more on the 38th day.
c. Sample answer: It is unreasonable because there
are only so many hours in the day that can be
dedicated to rowing.
ANSWER: a. 30 minutes and 30 seconds
ANSWER: 74
32. CCSS SENSE-MAKING Mario began an exercise
program to get back in shape. He plans to row 5
minutes on his rowing machine the first day and
increase his rowing time by one minute and thirty
seconds each day.
a. How long will he row on the 18th day?
b. On what day will Mario first row an hour or
more?
c. Is it reasonable for this pattern to continue
indefinitely? Explain.
b. on the 38th day
c. Sample answer: It is unreasonable because there
are only so many hours in the day that can be
dedicated to rowing.
Determine whether each sequence is
geometric. Write yes or no.
33. 21, 14, 7, …
SOLUTION: Find the ratio of the consecutive terms.
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SOLUTION: a. Given a 1 = 5, d = 1.5 and n = 18.
Page 13
Since the ratios are not the same, the sequence is not
geometric.
Since the ratios are the same, the sequence is
geometric.
c. Sample answer: It is unreasonable because there
are only so many hours in the day that can be
to rowing.
10-1dedicated
Sequences
as Functions
ANSWER: Yes
Determine whether each sequence is
geometric. Write yes or no.
36. 162, 108, 72, …
SOLUTION: Find the ratio of the consecutive terms.
33. 21, 14, 7, …
SOLUTION: Find the ratio of the consecutive terms.
Since the ratios are same, the sequence is geometric.
ANSWER: Yes
Since the ratios are not the same, the sequence is not
geometric.
ANSWER: No
34. 124, 186, 248, …
37. SOLUTION: Find the ratio of the consecutive terms.
SOLUTION: Find the ratio of the consecutive terms.
Since the ratios are not the same, the sequence is not
geometric.
ANSWER: No
35. –27, 18, –12, …
SOLUTION: Find the ratio of the consecutive terms.
Since the ratios are not same, the sequence is not
geometric.
ANSWER: No
38. –4, –2, 0, 2, …
Since the ratios are the same, the sequence is
geometric.
SOLUTION: Find the ratio of the consecutive terms.
ANSWER: Yes
36. 162, 108, 72, …
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eSolutions
SOLUTION: Find the ratio of the consecutive terms.
Since the ratios are not same, the sequence is not
Page 14
geometric.
ANSWER: 10-1No
Sequences as Functions
38. –4, –2, 0, 2, …
SOLUTION: Find the ratio of the consecutive terms.
ANSWER: No
Find the next three terms of the sequence.
Then graph the sequence.
39. 0.125, –0.5, 2, …
SOLUTION: Find the ratio of the consecutive terms.
Since the ratios are not same, the sequence is not
geometric.
ANSWER: No
Since the ratios are same, the sequence is geometric
Find the next three terms of the sequence.
Then graph the sequence.
To find the next term, multiply the previous term with
−4.
39. 0.125, –0.5, 2, …
SOLUTION: Find the ratio of the consecutive terms.
Graph the sequence.
Since the ratios are same, the sequence is geometric
To find the next term, multiply the previous term with
−4.
ANSWER: – 8, 32, –128
Graph the sequence.
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ANSWER: – 8, 32, –128
40. 18, 12, 8, …
SOLUTION: Page 15
ANSWER: 10-1 Sequences as Functions
40. 18, 12, 8, …
SOLUTION: Find the ratio of the consecutive terms.
Since the ratios are same, the sequence is geometric.
To find the next term, multiply the previous term by
.
41. 64, 48, 36, …
SOLUTION: Find the ratio of the consecutive terms.
Since the ratios are same, the sequence is geometric.
Graph the sequence.
To find the next term, multiply the previous term by
.
Graph the sequence.
ANSWER: ANSWER: eSolutions Manual - Powered by Cognero
41. 64, 48, 36, …
Page 16
10-1 Sequences as Functions
ANSWER: ANSWER: 42. 81, 108, 144, …
43. SOLUTION: Find the ratio of the consecutive terms.
SOLUTION: Find the ratio of the consecutive terms.
Since the ratios are same, the sequence is geometric
To find the next term, multiply the previous term by
.
Since the ratios are same, the sequence is geometric
To find the next term, multiply the previous term by
3.
Graph the sequence.
Graph the sequence.
ANSWER: eSolutions Manual - Powered by Cognero
ANSWER: 27, 81, 243
Page 17
10-1 Sequences as Functions
44. 1, 0.1, 0.01, 0.001, …
43. SOLUTION: Find the ratio of the consecutive terms.
SOLUTION: Find the ratio of the consecutive terms.
Since the ratios are same, the sequence is geometric.
Since the ratios are same, the sequence is geometric
To find the next term, multiply the previous term by
0.1.
To find the next term, multiply the previous term by
3.
Graph the sequence.
Graph the sequence.
ANSWER: 0.0001, 0.00001, 0.000001
ANSWER: 27, 81, 243
44. 1, 0.1, 0.01, 0.001, …
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SOLUTION: Find the ratio of the consecutive terms.
Determine whether each sequence is
arithmetic, geometric, or neither. Explain your
reasoning.
45. 3, 12, 27, 48, …
Page 18
10-1 Sequences as Functions
Determine whether each sequence is
arithmetic, geometric, or neither. Explain your
reasoning.
45. 3, 12, 27, 48, …
SOLUTION: Subtract each term from the term directly after it.
ANSWER: Neither; there is no common difference or ratio.
46. 1, –2, –5, –8, …
SOLUTION: Subtract each term from the term directly after it.
The common difference is –3.
Therefore, the sequence is arithmetic.
There is no common difference.
To find the common ratio, find the ratio of the
consecutive terms.
Therefore, the sequence is not arithmetic.
To find the common ratio, find the ratio of the
consecutive terms.
Since the ratios are not same, the sequence is not
geometric.
Since the ratios are not same, the sequence is not
geometric.
ANSWER: Neither; there is no common difference or ratio.
ANSWER: Arithmetic; the common difference is –3.
47. 12, 36, 108, 324, …
SOLUTION: Subtract each term from the term directly after it.
46. 1, –2, –5, –8, …
SOLUTION: Subtract each term from the term directly after it.
There is no common difference.
Therefore, this sequence is not arithmetic.
The common difference is –3.
Therefore, the sequence is arithmetic.
To find the common ratio, find the ratio of the
consecutive terms.
To find the common ratio, find the ratio of the
consecutive terms.
The common ratio is 3.
Since the ratios are same, the sequence is geometric.
Since the ratios are not same, the sequence is not
eSolutions Manual - Powered by Cognero
geometric.
ANSWER: Geometric; the common ratio is 3.
Page 19
ANSWER: Geometric; the common ratio is 3.
ANSWER: difference is –3.
10-1Arithmetic;
Sequencesthe
as common
Functions
47. 12, 36, 108, 324, …
48. SOLUTION: Subtract each term from the term directly after it.
SOLUTION: Subtract each term from the term directly after it.
There is no common difference.
Therefore, this sequence is not arithmetic.
There is no common difference.
Therefore, this sequence is not arithmetic.
To find the common ratio, find the ratio of the
consecutive terms.
To find the common ratio, find the ratio of the
consecutive terms.
The common ratio is 3.
Since the ratios are same, the sequence is geometric.
The common ratio is
ANSWER: Geometric; the common ratio is 3.
.
Since the ratios are the same, the sequence is
geometric.
ANSWER: 48. Geometric; the common ratio is
SOLUTION: Subtract each term from the term directly after it.
.
49. There is no common difference.
Therefore, this sequence is not arithmetic.
SOLUTION: Subtract each term from the term directly after it.
To find the common ratio, find the ratio of the
consecutive terms.
The common difference is
The common ratio is .
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Since the ratios are the same, the sequence is
.
Therefore, the sequence is arithmetic.
To find the common ratio, find the ratio of the
consecutive terms.
Page 20
ANSWER: ANSWER: Geometric; the common ratio is
10-1 Sequences as Functions
.
Arithmetic; the common difference is .
50. 6, 9, 14, 21, …
49. SOLUTION: Subtract each term from the term directly after it.
SOLUTION: Subtract each term from the term directly after it.
There is no common difference.
Therefore, the sequence is not arithmetic.
The common difference is
.
Therefore, the sequence is arithmetic.
To find the common ratio, find the ratio of the
consecutive terms.
To find the common ratio, find the ratio of the
consecutive terms.
Since the ratios are not same, the sequence is not
geometric.
Since the ratios are not same, the sequence is not
geometric.
ANSWER: Arithmetic; the common difference is .
50. 6, 9, 14, 21, …
SOLUTION: Subtract each term from the term directly after it.
ANSWER: Neither; there is no common difference or ratio.
51. READING Sareeta took an 800-page book on
vacation. If she was already on page 112 and is
going to be on vacation for 8 days, what is the
minimum number of pages she needs to read per day
to finish the book by the end of her vacation?
SOLUTION: The number of pages to be read is 800 – 112 or 688.
The minimum number of pages to read per day is
.
There is no common difference.
Therefore, the sequence is not arithmetic.
ANSWER: 86 pg/day
To find the common ratio, find the ratio of the
consecutive terms.
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eSolutions
Since the ratios are not same, the sequence is not
geometric.
52. DEPRECIATION Tammy’s car is expected to
depreciate at a rate of 15% per year. If her car is
currently valued at $24,000, to the nearest dollar,
how much will it be worth in 6 years?
Page 21
SOLUTION: Substitute 0.15, 6 and 24000 for r, t and P in
ANSWER: pg/day as Functions
10-186
Sequences
ANSWER: about 13,744 km
52. DEPRECIATION Tammy’s car is expected to
depreciate at a rate of 15% per year. If her car is
currently valued at $24,000, to the nearest dollar,
how much will it be worth in 6 years?
SOLUTION: Substitute 0.15, 6 and 24000 for r, t and P in
then evaluate.
The worth of the car will be about $9052 after 6
years.
ANSWER: $9052
53. CCSS REGULARITY When a piece of paper is
folded onto itself, it doubles in thickness. If a piece of
paper that is 0.1 mm thick could be folded 37 times,
how thick would it be?
54. REASONING Explain why the sequence 8, 10, 13,
17, 22 is not arithmetic.
SOLUTION: Sample answer: The consecutive terms do not share
a common difference. For instance, 22 – 17 = 5,
while 17 – 13 = 4.
ANSWER: Sample answer: The consecutive terms do not share
a common difference. For instance, 22 – 17 = 5,
while 17 – 13 = 4.
55. OPEN ENDED Describe a real-life situation that
can be represented by an arithmetic sequence with a
common difference of 8.
SOLUTION: Sample answer: A babysitter earns $20 for cleaning
the house and $8 extra for every hour she watches
the children.
SOLUTION: Given a 0 = 0.1, n = 37 and r = 2.
ANSWER: Sample answer: A babysitter earns $20 for cleaning
the house and $8 extra for every hour she watches
the children.
Find a 37.
The thickness would be about 13,744 km.
56. CHALLENGE The sum of three consecutive terms
of an arithmetic sequence is 6. The product of the
terms is –42. Find the terms.
SOLUTION: Let x be the first term in the arithmetic sequence.
Therefore, the next two terms should be x + d and x
+ 2d.
ANSWER: about 13,744 km
54. REASONING Explain why the sequence 8, 10, 13,
17, 22 is not arithmetic.
SOLUTION: Sample answer: The consecutive terms do not share
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a common difference. For instance, 22 – 17 = 5,
while 17 – 13 = 4.
Sample answer:
Let d = 5.
Therefore, the terms are –3, 2, 7.
Page 22
ANSWER: Sample answer: A babysitter earns $20 for cleaning
the house and $8 extra for every hour she watches
children. as Functions
10-1the
Sequences
56. CHALLENGE The sum of three consecutive terms
of an arithmetic sequence is 6. The product of the
terms is –42. Find the terms.
SOLUTION: Let x be the first term in the arithmetic sequence.
Therefore, the next two terms should be x + d and x
+ 2d.
Sample answer:
Let d = 5.
Therefore, the terms are –3, 2, 7.
ANSWER: –3, 2, 7
57. ERROR ANALYSIS Brody and Gen are
determining whether the sequence 8, 8, 8,… is
arithmetic, geometric, neither, or both. Is either of
them correct? Explain your reasoning.
Let d = 5.
Therefore, the terms are –3, 2, 7.
ANSWER: –3, 2, 7
57. ERROR ANALYSIS Brody and Gen are
determining whether the sequence 8, 8, 8,… is
arithmetic, geometric, neither, or both. Is either of
them correct? Explain your reasoning.
SOLUTION: Sample answer: Neither; the sequence is both
arithmetic and geometric.
ANSWER: Sample answer: Neither; the sequence is both
arithmetic and geometric.
58. OPEN ENDED Find a geometric sequence, an
arithmetic sequence, and a sequence that is neither
geometric nor arithmetic that begins 3, 9,… .
SOLUTION: Sample answer: geometric: 3, 9, 27, 81, 243, … arithmetic: 3, 9, 15, 21, 27, … neither geometric nor
arithmetic: 3, 9, 21, 45, 93, ...
SOLUTION: Sample answer: Neither; the sequence is both
arithmetic and geometric.
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eSolutions
ANSWER: Sample answer: Neither; the sequence is both
ANSWER: Sample answer: geometric: 3, 9, 27, 81, 243, … arithmetic: 3, 9, 15, 21, 27, … neither geometric nor
arithmetic: 3, 9, 21, 45, 93, ...
59. REASONING If a geometric sequence has a ratio r
such that
, what happens to the terms as n
increases? What would happen to the terms if
Page 23
?
ANSWER: Sample answer: geometric: 3, 9, 27, 81, 243, … arithmetic: 3, 9, 15, 21, 27, … neither geometric nor
3, as
9, 21,
45, 93, ...
10-1arithmetic:
Sequences
Functions
fraction. When
, the absolute value of the
terms will increase and approach infinity because
they are continuously being multiplied by a value
greater than 1.
59. REASONING If a geometric sequence has a ratio r
such that
, what happens to the terms as n
increases? What would happen to the terms if
?
SOLUTION: Sample answer: If a geometric sequence has a ratio
r such that
, as n increases, the absolute value
of the terms will decrease and approach zero
because they are continuously being multiplied by a
fraction. When
, the absolute value of the
terms will increase and approach infinity because
they are continuously being multiplied by a value
greater than 1.
ANSWER: Sample answer: If a geometric sequence has a ratio
r such that
, as n increases, the absolute value
of the terms will decrease and approach zero
because they are continuously being multiplied by a
fraction. When
, the absolute value of the
terms will increase and approach infinity because
they are continuously being multiplied by a value
greater than 1.
60. WRITING IN MATH Describe what happens to
the terms of a geometric sequence when the
common ratio is doubled. What happens when it is
halved? Explain your reasoning.
60. WRITING IN MATH Describe what happens to
the terms of a geometric sequence when the
common ratio is doubled. What happens when it is
halved? Explain your reasoning.
SOLUTION: Sample answer: When the value of r is doubled, a 2
doubles, a 3 quadruples, a 4 is multiplied by 8, a5 is
4
multiplied by 2 or 16, and so on. So, the new terms
are
. When the value of r is halved, the
new terms are
.
ANSWER: Sample answer: When the value of r is doubled, a 2
doubles, a 3 quadruples, a 4 is multiplied by 8, a5 is
4
multiplied by 2 or 16, and so on. So, the new terms
are
. When the value of r is halved, the
new terms are
.
61. SHORT RESPONSE Mrs. Aguilar’s rectangular
bedroom measures 13 feet by 11 feet. She wants to
purchase carpet for the bedroom that costs $2.95 per
square foot, including tax. How much will it cost to
carpet her bedroom?
SOLUTION: SOLUTION: Sample answer: When the value of r is doubled, a 2
doubles, a 3 quadruples, a 4 is multiplied by 8, a5 is
4
multiplied by 2 or 16, and so on. So, the new terms
are
. When the value of r is halved, the
new terms are
.
2
The area of the bedroom is 13 × 11 or 143 ft .
It costs $421.85 to carpet the bedroom.
ANSWER: $421.85
ANSWER: Sample answer: When the value of r is doubled, a 2
doubles, a 3 quadruples, a 4 is multiplied by 8, a5 is
4
multiplied by 2 or 16, and so on. So, the new terms
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are
. When the value of r is halved, the
new terms are
.
62. The pattern of filled circles and white circles below
can be described by a relationship between two
variables.
Page 24
Option C is the correct answer.
ANSWER: C
ANSWER: 10-1$421.85
Sequences as Functions
62. The pattern of filled circles and white circles below
can be described by a relationship between two
variables.
63. SAT/ACT Donna wanted to determine the average
of her six test scores. She added the scores correctly
to get T, but divided by 7 instead of 6. Her average
was 12 less than the actual average. Which equation
could be used to determine the value of T?
F G
Which rule relates w, the number of white circles, to
f, the number of dark circles?
A
H
J
B
K
C
SOLUTION: Donna’s average was 12 less than the actual
average.
D
SOLUTION: The relation between w and f is
Option C is the correct answer.
.
ANSWER: C
That is,
.
Rewrite the equation.
63. SAT/ACT Donna wanted to determine the average
of her six test scores. She added the scores correctly
to get T, but divided by 7 instead of 6. Her average
was 12 less than the actual average. Which equation
could be used to determine the value of T?
Option H is the correct answer.
ANSWER: H
F G
64. Find the next term in the geometric
sequence
H
A
J
K
eSolutions
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SOLUTION: Donna’s average was 12 less than the actual
average.
B
C
Page 25
ANSWER: 10-1HSequences as Functions
ANSWER: D
64. Find the next term in the geometric
Solve each system of equations.
sequence
65. A
SOLUTION: Substitute 5 for y in the quadratic equation and solve
for x.
B
C
D
The solutions are
SOLUTION: Find the common ratio.
ANSWER: .
66. The common ratio is
.
The next term in the geometric sequence is
.
SOLUTION: Solve the linear equation for y in terms of x.
Therefore, option D is the correct answer.
ANSWER: D
Substitute x + 1 for y in the quadratic equation and
solve for x.
Solve each system of equations.
65. SOLUTION: Substitute 5 for y in the quadratic equation and solve
for x.
By the Zero Product Property:
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Page 26
Substitute the values of x in the linear equation and
solve for y.
ANSWER: ANSWER: (–4, –3), (3, 4)
10-1 Sequences as Functions
66. 67. SOLUTION: Solve the linear equation for y in terms of x.
SOLUTION: 2
Substitute 3x for y in the second equation and solve
for x.
Substitute x + 1 for y in the quadratic equation and
solve for x.
Since the radicand is negative, there is no solution.
ANSWER: no solution
By the Zero Product Property:
Write each equation in standard form. State
whether the graph of the equation is a parabola,
circle, ellipse, or hyperbola. Then graph the
equation.
Substitute the values of x in the linear equation and
solve for y.
68. SOLUTION: Divide the equation by 6 on both sides.
The solutions are (–4, –3) and (3, 4).
ANSWER: (–4, –3), (3, 4)
The equation is in the standard form of circle.
67. SOLUTION: 2
Substitute 3x for y in the second equation and solve
for x.
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ANSWER: Page 27
ANSWER: solution as Functions
10-1no
Sequences
Write each equation in standard form. State
whether the graph of the equation is a parabola,
circle, ellipse, or hyperbola. Then graph the
equation.
69. SOLUTION: 68. SOLUTION: Divide the equation by 6 on both sides.
The equation is in the standard form of hyperbola.
The equation is in the standard form of circle.
ANSWER: hyperbola;
ANSWER: 2
2
circle; x + y = 27
70. SOLUTION: 69. SOLUTION: The equation is in the standard form of circle.
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Page 28
10-1 Sequences as Functions
Graph each function.
70. 71. SOLUTION: SOLUTION: The equation is in the standard form of circle.
The vertical asymptotes are at x = 2 and x = –3.
Since the degree of the numerator is less than the
degree of the denominator, the horizontal asymptote
is at y = 0.
ANSWER: 2
2
circle; x + (y + 3) = 36
ANSWER: Graph each function.
71. SOLUTION: 72. SOLUTION: The vertical asymptotes are at x = 2 and x = –3.
Since the degree of the numerator is less than the
degree of the denominator, the horizontal asymptote
is at y = 0.
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The vertical asymptote is at x = 2.
Page 29
Since the degree of the numerator is less than the
10-1 Sequences as Functions
72. 73. SOLUTION: SOLUTION: The vertical asymptote is at x = 2.
Since the degree of the numerator is less than the
degree of the denominator, the horizontal asymptote
is at y = 0.
The graph of
ANSWER: ANSWER: is same as the graph of
f(x) = x – 6.
74. HEALTH A certain medication is eliminated from
the bloodstream at a steady rate. It decays according
–0.1625t
to the equation y = ae
, where t is in hours.
Find the half-life of this substance.
73. SOLUTION: SOLUTION: Substitute 0.5a for y and solve for t.
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eSolutions
The graph of
Page 30
is same as the graph of
10-1 Sequences as Functions
74. HEALTH A certain medication is eliminated from
the bloodstream at a steady rate. It decays according
–0.1625t
to the equation y = ae
, where t is in hours.
Find the half-life of this substance.
SOLUTION: Substitute 0.5a for y and solve for t.
ANSWER: y = 0.5x + 1
76. passes through
SOLUTION: Substitute
for m, x1 and y 1 in the
point-slope form of a line.
The half-life of the substance is about 4.27 hours.
ANSWER: about 4.27 hours
ANSWER: Write an equation of each line.
75. passes through (6, 4), m = 0.5
77. passes through (0, –6), m = 3
SOLUTION: Substitute 0.5, 6 and 4 for m, x1 and y 1 in the pointslope form of a line.
SOLUTION: Substitute 3, 0 and –6 for m, x1 and y 1 in the pointslope form of a line.
ANSWER: y = 3x – 6
ANSWER: y = 0.5x + 1
78. passes through (0, 4),
76. passes through
SOLUTION: SOLUTION: eSolutions
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Substitute
, 0 and 4 for m, x1 and y 1 in the point-
Page 31
slope form of a line.
Substitute
for m, x1 and y 1 in the
ANSWER: ANSWER: = 3x – 6 as Functions
10-1ySequences
78. passes through (0, 4),
79. passes through (1, 3) and
SOLUTION: SOLUTION: Find the slope of the line.
Substitute
, 0 and 4 for m, x1 and y 1 in the point-
slope form of a line.
ANSWER: Substitute
, 1 and 3 for m, x1 and y 1 in the point-
slope form of a line.
79. passes through (1, 3) and
SOLUTION: Find the slope of the line.
ANSWER: 80. passes through (–5, 1) and (5, 16)
Substitute
, 1 and 3 for m, x1 and y 1 in the point-
slope form of a line.
SOLUTION: Find the slope of the line.
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Page 32
ANSWER: 10-1 Sequences as Functions
80. passes through (–5, 1) and (5, 16)
SOLUTION: Find the slope of the line.
Substitute
, 5 and 16 for m, x1 and y 1 in the point-
slope form of a line.
ANSWER: eSolutions Manual - Powered by Cognero
Page 33
ANSWER: 233
10-2 Arithmetic Sequences and Series
Find the indicated term of each arithmetic
sequence.
Write an equation for the nth term of each
arithmetic sequence.
1. a 1 = 14, d = 9, n = 11
3. 13, 19, 25, …
SOLUTION: SOLUTION: 19 – 13 = 6
25 – 19 = 6
The common difference d is 6.
a 1 = 13
ANSWER: 104
2. a 18 for 12, 25, 38, …
SOLUTION: 25 – 12 = 13
38 – 25 = 13
ANSWER: a n = 6n + 7
The common difference d is 13.
a 1 = 12
4. a 5 = –12, d = –4
SOLUTION: ANSWER: 233
Use the value of a 1 to find the nth term.
Write an equation for the nth term of each
arithmetic sequence.
3. 13, 19, 25, …
SOLUTION: 19 – 13 = 6
25 – 19 = 6
ANSWER: a n = –4n + 8
The common difference d is 6.
Find the arithmetic means in each sequence.
a = 13
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eSolutions
Page 1
5. ANSWER: a n = –4n + 8Sequences and Series
10-2 Arithmetic
ANSWER: –1, 2, 5
Find the arithmetic means in each sequence.
Find the sum of each arithmetic series.
7. the first 50 natural numbers
5. SOLUTION: Here a 1 = 6 and a 5 = 42.
SOLUTION: Therefore, the missing numbers are (6 + 9) or 15,
(15 + 9) or 24, (24 + 9) or 33.
ANSWER: 1275
ANSWER: 15, 24, 33
6. 8. 4 + 8 + 12 + … + 200
SOLUTION: 8– 4=4
12 – 8 = 4
SOLUTION: Here a 1 = –4 and a 5 = 8.
Therefore, the missing numbers are (–4 + 3) or −1,
(–1 + 3) or 2, (2 + 3) or 5.
The common difference is 4.
Find the sum of the series.
ANSWER: –1, 2, 5
Find the sum of each arithmetic series.
7. the first 50 natural numbers
SOLUTION: ANSWER: 5100
9. a 1 = 12, a n = 188, d = 4
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Page 2
SOLUTION: ANSWER: 5100
10-2 Arithmetic
Sequences and Series
9. a 1 = 12, a n = 188, d = 4
SOLUTION: ANSWER: 1995
Find the first three terms of each arithmetic
series.
11. a 1 = 8, a n = 100, S n = 1296
SOLUTION: Find the sum of the series.
ANSWER: 4500
10. a n = 145, d = 5, n = 21
Therefore, the first three terms are 8, (8 + 4) or 12,
(12 + 4) or 16.
ANSWER: 8, 12, 16
SOLUTION: 12. n = 18, a n = 112, S n = 1098
SOLUTION: Find the sum of the series.
ANSWER: 1995
Find the first three terms of each arithmetic
series.
11. a 1 = 8, a n = 100, S n = 1296
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SOLUTION: Therefore, the first three terms are 10, (10 + 6) or
16, (16 + 6) or 22.
ANSWER: 10, 16, 22
Page 3
ANSWER: 8, 12, 16
10-2 Arithmetic
Sequences and Series
12. n = 18, a n = 112, S n = 1098
SOLUTION: ANSWER: 10, 16, 22
13. MULTIPLE CHOICE Find
A 45
B 78
C 342
D 410
SOLUTION: There are 12 – 1 + 1 or 12 terms, so n = 12.
Therefore, the first three terms are 10, (10 + 6) or
16, (16 + 6) or 22.
Find the sum.
ANSWER: 10, 16, 22
13. MULTIPLE CHOICE Find
A 45
B 78
C 342
Option C is the correct answer.
ANSWER: C
Find the indicated term of each arithmetic
sequence.
D 410
SOLUTION: There are 12 – 1 + 1 or 12 terms, so n = 12.
Find the sum.
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14. a 1 = –18, d = 12, n = 16
SOLUTION: ANSWER: 162
15. a 1 = –12, n = 66, d = 4
Page 4
ANSWER: C
10-2 Arithmetic
Sequences and Series
Find the indicated term of each arithmetic
sequence.
ANSWER: –129
17. a 15 for –5, –12, –19, …
SOLUTION: –12 – (–5) = –7
–19 – (–12) = –7
SOLUTION: 14. a 1 = –18, d = 12, n = 16
The common difference d is –7.
a 1 = –5
ANSWER: 162
15. a 1 = –12, n = 66, d = 4
ANSWER: –103
SOLUTION: 18. a 10 for –1, 1, 3, …
ANSWER: 248
16. a 1 = 9, n = 24, d = –6
SOLUTION: 1 – (–1) = 2
3– 1=2
The common difference d is 2.
a 1 = –1
SOLUTION: ANSWER: –129
ANSWER: 17
19. a 24 for 8.25, 8.5, 8.75, …
17. a 15 for –5, –12, –19, …
SOLUTION: –12 – (–5) = –7
–19 – (–12) = –7
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SOLUTION: 8.5 – 8.25 = 0.25
8.75 – 8.5 = 0.25
The common difference d is 0.25.
The common difference d is –7.
a 1 = 8.25
Page 5
ANSWER: 17
10-2 Arithmetic
Sequences and Series
19. a 24 for 8.25, 8.5, 8.75, …
ANSWER: a n = 11n + 13
21. 31, 17, 3, …
SOLUTION: 8.5 – 8.25 = 0.25
8.75 – 8.5 = 0.25
SOLUTION: 17 – 31 = –14
3 – 17 = –14
The common difference d is 0.25.
a 1 = 8.25
ANSWER: 14
Write an equation for the nth term of each
arithmetic sequence.
20. 24, 35, 46, …
The common difference d is –14.
a 1 = 31
ANSWER: a n = –14n + 45
22. a 9 = 45, d = –3
SOLUTION: SOLUTION: 35 – 24 = 11
46 – 35 = 11
The common difference d is 11.
a 1 = 24
Use the value of a 1 to find the nth term.
ANSWER: a n = 11n + 13
21. 31, 17, 3, …
ANSWER: a n = –3n + 72
23. a 7 = 21, d = 5
SOLUTION: SOLUTION: eSolutions
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17 Manual
– 31 =- –14
3 – 17 = –14
Page 6
ANSWER: a n = –3n + 72
10-2 Arithmetic
Sequences and Series
23. a 7 = 21, d = 5
ANSWER: a n = 0.25n + 11
25. a 5 = 1.5, d = 4.5
SOLUTION: SOLUTION: Use the value of a 1 to find the nth term.
Use the value of a 1 to find the nth term.
ANSWER: a n = 5n – 14
ANSWER: a n = 4.5n – 21
24. a 4 = 12, d = 0.25
26. 9, 2, –5, …
SOLUTION: SOLUTION: 2 – 9 = –7
–5 – 2 = –7
The common difference d is –7.
Use the value of a 1 to find the nth term.
a1 = 9
ANSWER: a n = 0.25n + 11
25. a 5 = 1.5, d = 4.5
ANSWER: a n = –7n + 16
27. a 6 = 22, d = 9
SOLUTION: SOLUTION: eSolutions Manual - Powered by Cognero
Page 7
ANSWER: a n = –7n + 16
10-2 Arithmetic
Sequences and Series
ANSWER: a n = –2n + 8
27. a 6 = 22, d = 9
29. SOLUTION: SOLUTION: Use the value of a 1 to find the nth term.
Use the value of a 1 to find the nth term.
ANSWER: a n = 9n – 32
28. a 8 = –8, d = –2
ANSWER: SOLUTION: 30. –12, –17, –22, …
SOLUTION: –17 – (–12) = –5
–22 – (–17) = –5
Use the value of a 1 to find the nth term.
The common difference d is –5.
a 1 = –12
ANSWER: a n = –2n + 8
29. ANSWER: a n = –5n – 7
SOLUTION: eSolutions Manual - Powered by Cognero
31. Page 8
ANSWER: ANSWER: a n = –5n – 7
10-2 Arithmetic Sequences and Series
30. –12, –17, –22, …
SOLUTION: –17 – (–12) = –5
–22 – (–17) = –5
31. SOLUTION: The common difference d is –5.
a 1 = –12
Use the value of a 1 to find the nth term.
ANSWER: a n = –5n – 7
31. SOLUTION: ANSWER: 32. CCSS STRUCTURE José averaged 123 total pins per game in his bowing league this season. He
is taking bowling lessons and hopes to bring his
average up by 8 pins each new season.
Use the value of a 1 to find the nth term.
a. Write an equation to represent the nth term of
the sequence.
b. If the pattern continues, during what season will
José average 187 per game?
c. Is it reasonable for this pattern to continue
indefinitely? Explain.
ANSWER: SOLUTION: a. Given d = 8 and a 1 = 123.
Find the nth term.
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32. CCSS STRUCTURE José averaged 123 total pins per game in his bowing league this season. He
Page 9
ANSWER: c. Sample answer: No; there are a maximum of 300
points in a bowling game, so it would be impossible
for the average to continue to climb indefinitely.
10-2 Arithmetic Sequences and Series
32. CCSS STRUCTURE José averaged 123 total pins per game in his bowing league this season. He
is taking bowling lessons and hopes to bring his
average up by 8 pins each new season.
Find the arithmetic means in each sequence.
33. SOLUTION: Here a 1 = 24 and a 6 = –1.
a. Write an equation to represent the nth term of
the sequence.
b. If the pattern continues, during what season will
José average 187 per game?
c. Is it reasonable for this pattern to continue
indefinitely? Explain.
SOLUTION: a. Given d = 8 and a 1 = 123.
Find the nth term.
Therefore, the missing numbers are (24 – 5) or 19,
(19 – 5) or 14, (14 – 5) or 9, and (9 – 5) or 4.
ANSWER: 19, 14, 9, 4
b. Substitute 187 for a n and solve for n.
34. ]
SOLUTION: Here a 1 = –6 and a 6 = 49.
Therefore José’s average will be 187 pins per game
th
in the 9 season.
c. Sample answer: No; there are a maximum of 300
points in a bowling game, so it would be impossible
for the average to continue to climb indefinitely.
Therefore, the missing numbers are (–6 + 11) or 5,
(5 + 11) or 16, (16 + 11) or 27, and (27 + 11) or 38.
ANSWER: a. a n = 115 + 8n
ANSWER: 5, 16, 27, 38
b. 9th season
c. Sample answer: No; there are a maximum of 300
points in a bowling game, so it would be impossible
for the average to continue to climb indefinitely.
35. Find the arithmetic means in each sequence.
SOLUTION: Here a 1 = –28 and a 6 = 7.
33. eSolutions Manual - Powered by Cognero
SOLUTION: Here a 1 = 24 and a 6 = –1.
Page 10
ANSWER: 5, 16, 27, 38 Sequences and Series
10-2 Arithmetic
ANSWER: 75, 66, 57, 48
35. 37. SOLUTION: Here a 1 = –28 and a 6 = 7.
SOLUTION: Here a 1 = –12 and a 7 = –66.
Therefore, the missing numbers are (–28 + 7) or –
21, (–21 + 7) or –14, (–14 + 7) or –7, and (–7 + 7)
or 0.
Therefore, the missing numbers are (–12 – 9) or –
21, (–21 – 9) or –30, (–30 – 9) or –39, (–39 – 9) or
–48, and (–48 – 9) or –57.
ANSWER: –21, –14, –7, 0
ANSWER: –21, –30, –39, –48, –57
36. 38. SOLUTION: Here a 1 = 84 and a 6 = 39.
SOLUTION: Here a 1 = 182 and a 7 = 104.
Therefore, the missing numbers are (84 – 9) or 75,
(75 – 9) or 66, (66 – 9) or 57, and (57 – 9) or 48.
Therefore, the missing numbers are (182 – 13) or
169, (169 – 13) or 156, (156 – 13) or 143, (143 –
13) or 130, and (130 – 13) or 117.
ANSWER: 75, 66, 57, 48
ANSWER: 169, 156, 143, 130, 117
37. SOLUTION: Here a 1 = –12 and a 7 = –66.
Find the sum of each arithmetic series.
39. the first 100 even natural numbers
SOLUTION: Here a 1 = 2 and a 100 = 200.
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Find the sum.
Page 11
ANSWER: 169, 156, 143,Sequences
130, 117 and Series
10-2 Arithmetic
Find the sum of each arithmetic series.
39. the first 100 even natural numbers
SOLUTION: Here a 1 = 2 and a 100 = 200.
ANSWER: 40,000
41. the first 100 odd natural numbers
SOLUTION: Here a 1 = 1 and a 100 = 199.
n = 100
Find the sum.
Find the sum.
ANSWER: 10,100
ANSWER: 10,000
40. the first 200 odd natural numbers
SOLUTION: Here a 1 = 1 and a 200 = 399.
Find the sum.
42. the first 300 even natural numbers
SOLUTION: Here a 1 = 2 and a 300 = 300.
n = 300
Find the sum.
ANSWER: 40,000
41. the first 100 odd natural numbers
SOLUTION: Here a 1 = 1 and a 100 = 199.
ANSWER: 90,300
43. –18 + (–15) + (–12) + … + 66
n = 100
SOLUTION: –15 – (–18) = 3
–12 – (–15) = 3
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Find the sum.
The common difference is 3.
Page 12
ANSWER: 90,300
10-2 Arithmetic
Sequences and Series
43. –18 + (–15) + (–12) + … + 66
ANSWER: 696
44. –24 + (–18) + (–12) + … + 72
SOLUTION: –15 – (–18) = 3
–12 – (–15) = 3
SOLUTION: –18 – (–24) = 6
–12 – (–18) = 6
The common difference is 3.
The common difference is 6.
Find the value of n.
Find the value of n.
Find the sum.
Find the sum.
ANSWER: 696
ANSWER: 408
44. –24 + (–18) + (–12) + … + 72
SOLUTION: –18 – (–24) = 6
–12 – (–18) = 6
45. a 1 = –16, d = 6, n = 24
SOLUTION: The common difference is 6.
Find the value of n.
ANSWER: 1272
46. n = 19, a n = 154, d = 8
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SOLUTION: Page 13
ANSWER: 1272
10-2 Arithmetic
Sequences and Series
46. n = 19, a n = 154, d = 8
SOLUTION: Find the value of a 1.
ANSWER: 1558
47. CONTESTS The prizes in a weekly radio contest
began at $150 and increased by $50 for each week
that the contest lasted. If the contest lasted for
eleven weeks, how much was awarded in total?
SOLUTION: Given, a 1 = 150, d = 50 and n = 11.
Find the value of a 11.
Find the sum.
Find the sum.
ANSWER: 1558
47. CONTESTS The prizes in a weekly radio contest
began at $150 and increased by $50 for each week
that the contest lasted. If the contest lasted for
eleven weeks, how much was awarded in total?
A cash prizes totaled $4400 for the eleven week
contest.
SOLUTION: Given, a 1 = 150, d = 50 and n = 11.
Find the value of a 11.
ANSWER: $4400
Find the first three terms of each arithmetic
series.
48. n = 32, a n = –86, S n = 224
Find the sum.
SOLUTION: Find the value of a 1.
A cash prizes totaled $4400 for the eleven week
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contest.
Page 14
Find the value of d.
ANSWER: $4400
10-2 Arithmetic
Sequences and Series
Find the first three terms of each arithmetic
series.
48. n = 32, a n = –86, S n = 224
ANSWER: 100, 94, 88
49. a 1 = 48, a n = 180, S n = 1368
SOLUTION: Find the value of n.
SOLUTION: Find the value of a 1.
Find the value of d.
Find the value of d.
So, the sequence is 48, 60, 72, …
ANSWER: 48, 60, 72
So, the sequence is 100, 94, 88, …
50. a 1 = 3, a n = 66, S n = 759
ANSWER: 100, 94, 88
SOLUTION: Find the value of n.
49. a 1 = 48, a n = 180, S n = 1368
SOLUTION: Find the value of n.
Find the value of d.
Find the value of d.
Therefore, the first three terms are 3, 6 and 9.
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ANSWER: 3, 6, 9
Page 15
ANSWER: 48, 60, 72 Sequences and Series
10-2 Arithmetic
50. a 1 = 3, a n = 66, S n = 759
ANSWER: 3, 6, 9
51. n = 28, a n = 228, S n = 2982
SOLUTION: Find the value of n.
SOLUTION: Find the value of a 1.
Find the value of d.
Find the value of d.
Therefore, the first three terms are 3, 6 and 9.
ANSWER: 3, 6, 9
51. n = 28, a n = 228, S n = 2982
SOLUTION: Find the value of a 1.
Therefore, the first three terms are –15, –6 and 3.
ANSWER: –15, –6, 3
52. a 1 = –72, a n = 453, S n = 6858
SOLUTION: Find the value of n.
Find the value of d.
Find the value of d.
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Therefore, the first three terms are –15, –6 and 3.
Therefore, the first three terms are –72, –57 and –
42.
ANSWER: –15, –6, 3 Sequences and Series
10-2 Arithmetic
52. a 1 = –72, a n = 453, S n = 6858
ANSWER: –72, –57, –42
53. n = 30, a n = 362, S n = 4770
SOLUTION: Find the value of n.
SOLUTION: Find the value of a 1.
Find the value of d.
Find the value of d.
Therefore, the first three terms are –72, –57 and –
42.
Therefore, the first three terms are –44, –30 and –
16.
ANSWER: –72, –57, –42
ANSWER: –44, –30, –16
53. n = 30, a n = 362, S n = 4770
SOLUTION: Find the value of a 1.
54. a 1 = 19, n = 44, S n = 9350
SOLUTION: Find the value of a n.
Find the value of d.
Find the value of d.
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Therefore, the first three terms are –44, –30 and –
16.
Page 17
ANSWER: –44, –30, –16Sequences and Series
10-2 Arithmetic
54. a 1 = 19, n = 44, S n = 9350
ANSWER: 19, 28, 37
55. a 1 = –33, n = 36, S n = 6372
SOLUTION: Find the value of a n.
SOLUTION: Find the value of a n.
Find the value of d.
Find the value of d.
Therefore, the first three terms are 19, 28 and 37.
Therefore, the first three terms are –33, –21 and –
9.
ANSWER: 19, 28, 37
ANSWER: –33, –21, –9
55. a 1 = –33, n = 36, S n = 6372
SOLUTION: Find the value of a n.
56. PRIZES A radio station is offering a total of $8500
in prizes over ten hours. Each hour, the prize will
increase by $100. Find the amounts of the first and
last prize.
SOLUTION: Given n = 10, d = 100 and S 10 = 8500.
Find the value of a 1.
Find the value of d.
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Find the value of a 10.
Page 18
ANSWER: –33, –21, –9Sequences and Series
10-2 Arithmetic
ANSWER: $400 and $1300
56. PRIZES A radio station is offering a total of $8500
in prizes over ten hours. Each hour, the prize will
increase by $100. Find the amounts of the first and
last prize.
Find the sum of each arithmetic series.
57. SOLUTION: Given n = 10, d = 100 and S 10 = 8500.
SOLUTION: There are 16 – 1 + 1 or 16 terms, so n = 16.
Find the value of a 1.
Find the sum.
Find the value of a 10.
Therefore,
.
ANSWER: 512
ANSWER: $400 and $1300
Find the sum of each arithmetic series.
58. 57. SOLUTION: There are 16 – 1 + 1 or 16 terms, so n = 16.
SOLUTION: There are 13 – 4 + 1 or 10 terms, so n = 10.
Find the sum.
Find the sum.
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Therefore,
Page 19
.
ANSWER: 512
10-2 Arithmetic
Sequences and Series
58. ANSWER: 350
59. SOLUTION: There are 13 – 4 + 1 or 10 terms, so n = 10.
SOLUTION: There are 16 – 5 + 1 or 12 terms, so n = 12.
Find the sum.
Find the sum.
Therefore,
.
Therefore,
ANSWER: 350
ANSWER: 324
59. .
60. SOLUTION: There are 16 – 5 + 1 or 12 terms, so n = 12.
SOLUTION: There are 12 – 0 + 1 or 13 terms, so n = 13.
Find the sum.
Find the sum.
Therefore,
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.
Therefore,
ANSWER: ANSWER: .
Page 20
ANSWER: 324
10-2 Arithmetic
Sequences and Series
ANSWER: –208
60. SOLUTION: There are 12 – 0 + 1 or 13 terms, so n = 13.
61. FINANCIAL LITERACY Daniela borrowed
some money from her parents. She agreed to pay
$50 at the end of the first month and $25 more each
additional month for 12 months. How much does
she pay in total after the 12 months?
SOLUTION: Given a 1 = 50, d = 25 and n = 12.
Find the sum.
Find the sum.
She pays $2250.
Therefore,
.
ANSWER: $2250
ANSWER: –208
61. FINANCIAL LITERACY Daniela borrowed
some money from her parents. She agreed to pay
$50 at the end of the first month and $25 more each
additional month for 12 months. How much does
she pay in total after the 12 months?
SOLUTION: Given a 1 = 50, d = 25 and n = 12.
62. GRAVITY When an object is in free fall and air
resistance is ignored, it falls 16 feet in the first
second, an additional 48 feet during the next second,
and 80 feet during the third second. How many total
feet will the object fall in 10 seconds?
SOLUTION: Given a 1 = 16, n = 10.
The common difference d is 32.
Find the sum.
Find the sum.
The object will fall 1600 ft in 10 seconds.
She pays $2250.
ANSWER: $2250
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ANSWER: 1600 ft
Page 21
Use the given information to write an equation
ANSWER: $2250
10-2 Arithmetic
Sequences and Series
62. GRAVITY When an object is in free fall and air
resistance is ignored, it falls 16 feet in the first
second, an additional 48 feet during the next second,
and 80 feet during the third second. How many total
feet will the object fall in 10 seconds?
SOLUTION: Given a 1 = 16, n = 10.
ANSWER: 1600 ft
Use the given information to write an equation
that represents the nth term in each arithmetic
sequence.
63. The 100th term of the sequence is 245. The
common difference is 13.
SOLUTION: Given a 100 = 245, d = 13 and n = 100.
The common difference d is 32.
Find the value of a 1.
Find the sum.
The object will fall 1600 ft in 10 seconds.
Substitute the values of a 1 and d to find the nth
term.
ANSWER: 1600 ft
Use the given information to write an equation
that represents the nth term in each arithmetic
sequence.
ANSWER: a n = 13n – 1055
63. The 100th term of the sequence is 245. The
common difference is 13.
SOLUTION: Given a 100 = 245, d = 13 and n = 100.
Find the value of a 1.
64. The eleventh term of the sequence is 78. The
common difference is –9.
SOLUTION: Given a 11 = 78, d = –9 and n = 11.
Find the value of a 1.
Substitute the values of a 1 and d to find the nth
term.
Substitute the values of a 1 and d to find the nth
term.
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ANSWER: a n = 13n – 1055
10-2 Arithmetic
Sequences and Series
64. The eleventh term of the sequence is 78. The
common difference is –9.
ANSWER: a n = –9n + 177
65. The sixth term of the sequence is –34. The 23rd
term is 119.
SOLUTION: Given a 11 = 78, d = –9 and n = 11.
SOLUTION: Given a 6 = –34 and a 23 = 119.
Find the value of a 1.
Therefore, there are (23 – 6 + 1) or 18 terms
between –34 and 119.
Find the common difference of the series with a 1 =
–34 and a 18 = 119.
Substitute the values of a 1 and d to find the nth
term.
Find the value of a 1.
ANSWER: a n = –9n + 177
Substitute the values of a 1 and d to find the nth
65. The sixth term of the sequence is –34. The 23rd
term is 119.
term.
SOLUTION: Given a 6 = –34 and a 23 = 119.
Therefore, there are (23 – 6 + 1) or 18 terms
between –34 and 119.
Find the common difference of the series with a 1 =
–34 and a 18 = 119.
ANSWER: a n = 9n – 88
66. The 25th term of the sequence is 121. The 80th
term is 506.
SOLUTION: Given a 25 = 121 and a 80 = 506.
Find the value of a 1.
Therefore, there are (80 – 25 + 1) or 56 terms
between 121 and 506.
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Find the common difference of the series with a 1 =
Page 23
121 and a 56 = 506.
ANSWER: a n = 9n – 88Sequences and Series
10-2 Arithmetic
66. The 25th term of the sequence is 121. The 80th
term is 506.
SOLUTION: Given a 25 = 121 and a 80 = 506.
Therefore, there are (80 – 25 + 1) or 56 terms
between 121 and 506.
Find the common difference of the series with a 1 =
121 and a 56 = 506.
arrangements.
a. Make drawings to find the next three numbers as
tables are added one at a time to the arrangement.
b. Write an equation representing the nth number in
this pattern.
c. Is it possible to have seating for exactly 100
people with such an arrangement? Explain.
SOLUTION: a. For each increase in the number of table, the
number of people who can sit is increased by 4.
That is, the common difference is 4.
Therefore, the next three numbers are (10 + 4) or
14, (14 + 4) or 18 and (18 + 4) or 22.
Find the value of a 1.
Substitute the values of a 1 and d to find the nth
term.
b. Substitute a 1 = 6 and d = 4 in
ANSWER: a n = 7n – 54
.
c. No; there is no whole number n for which
.
67. CCSS MODELING The rectangular tables in a
reception hall are often placed end-to-end to form
one long table. The diagrams below show the
number of people who can sit at each of the table
arrangements.
ANSWER: a. 14, 18, 22
a. Make drawings to find the next three numbers as
to the arrangement.
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tables
are- added
at a time
Page 24
b. Write an equation representing the nth number in
c. No; there is no whole number n for which
.
ANSWER: 13
10-2 Arithmetic Sequences and Series
ANSWER: a. 14, 18, 22
69. SALARY Terry currently earns $28,000 per year.
If Terry expects a $4000 increase in salary every
year, after how many years will he have a salary of
$100,000 per year?
SOLUTION: Given a 1 = 28000, d = 4000 and a n = 100000.
Substitute the values of a 1, a n and d and solve for
n.
b. p n = 4n + 2
c. No; there is no whole number n for which 4n + 2
= 100.
68. PERFORMANCE A certain company pays its
employees according to their performance. Belinda
is paid a flat rate of $200 per week plus $24 for
every unit she completes. If she earned $512 in one
week, how many units did she complete?
SOLUTION: Given a 1 = 200, d = 24 and a n = 512.
Substitute the values of a 1, a n and d and solve for
n.
So he will have a salary of $100,000 per year after
the 19th year.
ANSWER: the 19th year
70. SPORTS While training for cross country, Silvia
plans to run 3 miles per day for the first week, and
then increase the distance by a half mile each of the
following weeks.
a. Write an equation to represent the nth term of
the sequence.
b. If the pattern continues, during which week will
she be running 10 miles per day?
14th term in the sequence is 512.
c. Is it reasonable for this pattern to continue
indefinitely? Explain.
Therefore, she completed (14 – 1) or 13 units.
SOLUTION: a. Given a 1 = 3 and d = 0.5.
ANSWER: 13
69. SALARY Terry currently earns $28,000 per year.
If Terry expects a $4000 increase in salary every
year, after how many years will he have a salary of
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SOLUTION: Page 25
b. 15th wk
ANSWER: the 19th yearSequences and Series
10-2 Arithmetic
c. Sample answer: No; eventually the number of
miles per day will become unrealistic.
70. SPORTS While training for cross country, Silvia
plans to run 3 miles per day for the first week, and
then increase the distance by a half mile each of the
following weeks.
a. Write an equation to represent the nth term of
the sequence.
71. MUTIPLE REPRESENTATIONS Consider
a. TABULAR Make a table of the partial sums of
the series for 1 ≤ k ≤ 10.
b. If the pattern continues, during which week will
she be running 10 miles per day?
c. Is it reasonable for this pattern to continue
indefinitely? Explain.
b. GRAPHICAL Graph (k, partial sum).
c. GRAPHICAL Graph f (x) = x2 + 3x on the
same grid.
d. VERBAL What do you notice about the two
graphs?
SOLUTION: a. Given a 1 = 3 and d = 0.5.
e. ANALYTICAL What conclusions can you
make about the relationship between quadratic
functions and the sum of arithmetic series?
Find the nth term.
f. ALGEBRAIC Find the arithmetic series that
2
relates to g(x) = x + 8x.
b. Substitute 10 for a n in
and solve SOLUTION: a.
for n.
During 15th week, she will be running 10 miles per
day.
c. Sample answer: No; eventually the number of
miles per day will become unrealistic.
b.
ANSWER: a. a n = 2.5 + 0.5n
b. 15th wk
c. Sample answer: No; eventually the number of
miles per day will become unrealistic.
71. MUTIPLE REPRESENTATIONS Consider
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a. TABULAR Make a table of the partial sums of
c.
Page 26
10-2 Arithmetic Sequences and Series
c.
c.
d. Sample answer: The graphs cover the same
range. The domain of the series is the natural
numbers, while the domain of the quadratic function
is all real numbers, 0 ≤ x ≤ 10.
e . Sample answer: For every partial sum of an
arithmetic series, there is a corresponding quadratic
function that shares the same range.
f.
f.
d. Sample answer: The graphs cover the same
range. The domain of the series is the natural
numbers, while the domain of the quadratic function
is all real numbers, 0 ≤ x ≤ 10.
e . Sample answer: For every partial sum of an
arithmetic series, there is a corresponding quadratic
function that shares the same range.
ANSWER: a.
Find the value of x.
72. SOLUTION: There are x – 3 + 1 or x – 2 terms, so n = x – 2.
b.
Find the sum.
c.
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Equate the sum with the given value and solve for
x.
Page 27
ANSWER: 18
f.
10-2 Arithmetic Sequences and Series
Find the value of x.
73. 72. SOLUTION: There are x – 3 + 1 or x – 2 terms, so n = x – 2.
SOLUTION: There are x – 5 + 1 or x – 4 terms, so n = x – 4.
Find the sum.
Find the sum.
Equate the sum with the given value and solve for
x.
Equate the sum with the given value and solve for
x.
The value of x should be positive. Therefore, x =
18.
ANSWER: 18
73. The value of x should be positive. Therefore, x =
16.
ANSWER: 16
74. CCSS CRITIQUE Eric and Juana are
determining the formula for the nth term for the
sequence –11, –2, 7, 16, … . Is either of them
correct? Explain your reasoning.
SOLUTION: eSolutions
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There are x – 5 + 1 or x – 4 terms, so n = x – 4.
Page 28
ANSWER: 16
10-2 Arithmetic
Sequences and Series
74. CCSS CRITIQUE Eric and Juana are
determining the formula for the nth term for the
sequence –11, –2, 7, 16, … . Is either of them
correct? Explain your reasoning.
ANSWER: Sample answer: Eric; Juana missed the step of
multiplying d by n – 1.
75. REASONING If a is the third term in an
arithmetic sequence, b is the fifth term, and c is the
eleventh term, express c in terms of a and b.
SOLUTION: Given a 3 = a, a 5 = b and a 11 = c.
Find the common difference.
Find the value of a 1.
SOLUTION: Sample answer: Eric; Juana missed the step of
multiplying d by n – 1.
Find the value of c in terms of a and b.
ANSWER: Sample answer: Eric; Juana missed the step of
multiplying d by n – 1.
75. REASONING If a is the third term in an
arithmetic sequence, b is the fifth term, and c is the
eleventh term, express c in terms of a and b.
SOLUTION: Given a 3 = a, a 5 = b and a 11 = c.
Find the common difference.
ANSWER: 4b – 3a
76. CHALLENGE There are three arithmetic means
between a and b in an arithmetic sequence. The
average of the arithmetic means is 16. What is the
average of a and b?
SOLUTION: The three arithmetic means between a and b are
.
Find the value of a 1.
The average of the arithmetic means is
.
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Therefore,
.
The term b can be written as a + 4d.
Page 29
ANSWER: 4b – 3a
10-2 Arithmetic
Sequences and Series
ANSWER: 16
76. CHALLENGE There are three arithmetic means
between a and b in an arithmetic sequence. The
average of the arithmetic means is 16. What is the
average of a and b?
SOLUTION: The three arithmetic means between a and b are
.
77. CHALLENGE Find S n for (x + y) + (x + 2y) + (x
+ 3y) + … .
SOLUTION: (x + 2y) – (x + y) = y
(x + 3y) – (x + 2y) = y
The common difference is y.
The average of the arithmetic means is
.
Find the sum.
Therefore,
.
The term b can be written as a + 4d.
The average of a and b is
.
We know that
.
Therefore, the average of a and b is 16.
ANSWER: 16
77. CHALLENGE Find S n for (x + y) + (x + 2y) + (x
+ 3y) + … .
SOLUTION: (x + 2y) – (x + y) = y
(x + 3y) – (x + 2y) = y
ANSWER: 78. OPEN ENDED Write an arithmetic series with 8
terms and a sum of 324.
The common difference is y.
SOLUTION: Sample answer: 9 + 18 + 27 + … + 72
ANSWER: Sample answer: 9 + 18 + 27 + … + 72
Find the sum.
79. WRITING IN MATH Compare and contrast
arithmetic sequences and series.
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SOLUTION: Sample answer: An arithmetic sequence is a list of
Page 30
terms such that any pair of successive terms has a
common difference. An arithmetic series is the sum
of the terms of an arithmetic sequence.
ANSWER: Sample answer:
9 + 18 +and
27 +Series
… + 72
10-2 Arithmetic
Sequences
Sample answer: An arithmetic sequence is a list of
terms such that any pair of successive terms has a
common difference. An arithmetic series is the sum
of the terms of an arithmetic sequence.
79. WRITING IN MATH Compare and contrast
arithmetic sequences and series.
80. PROOF Prove the formula for the nth term of an
arithmetic sequence.
SOLUTION: Sample answer: An arithmetic sequence is a list of
terms such that any pair of successive terms has a
common difference. An arithmetic series is the sum
of the terms of an arithmetic sequence.
SOLUTION: Sample answer:
Let a n = the nth term of the sequence and d = the
ANSWER: Sample answer: An arithmetic sequence is a list of
terms such that any pair of successive terms has a
common difference. An arithmetic series is the sum
of the terms of an arithmetic sequence.
common difference
a 2 = a 1 + d Definition of the second term of an
arithmetic sequence
a 3 = a 2 + d Definition of the third term of an
arithmetic sequence
a 3 = (a 1 + d ) + d Substitution
a 3 = a 1 + 2d Associative Property of Addition
a 3 = a 1 + (3 – 1)d 3 – 1 = 2
80. PROOF Prove the formula for the nth term of an
arithmetic sequence.
SOLUTION: Sample answer:
Let a n = the nth term of the sequence and d = the
common difference
a 2 = a 1 + d Definition of the second term of an
arithmetic sequence
a 3 = a 2 + d Definition of the third term of an
arithmetic sequence
a 3 = (a 1 + d ) + d Substitution
a 3 = a 1 + 2d Associative Property of Addition
a 3 = a 1 + (3 – 1)d 3 – 1 = 2
a n = a 1 + (n – 1)d n = 3
a n = a 1 + (n – 1)d n = 3
ANSWER: Sample answer:
Let a n = the nth term of the sequence and d = the
common difference
a 2 = a 1 + d Definition of the second term of an
arithmetic sequence
a 3 = a 2 + d Definition of the third term of an
arithmetic sequence
a 3 = (a 1 + d ) + d Substitution
a 3 = a 1 + 2d Associative Property of Addition
a 3 = a 1 + (3 – 1)d 3 – 1 = 2
a n = a 1 + (n – 1)d n = 3
ANSWER: Sample answer:
Let a n = the nth term of the sequence and d = the
common difference
a 2 = a 1 + d Definition of the second term of an
arithmetic sequence
a 3 = a 2 + d Definition of the third term of an
arithmetic sequence
a 3 = (a 1 + d ) + d Substitution
81. PROOF Derive a sum formula that does not
include a 1.
SOLUTION: General sum formula
a n = a 1 + (n – 1)d Formula for nth term
a 3 = a 1 + 2d Associative Property of Addition
a n – (n – 1)d = a 1 Subtract (n – 1)d from both
a 3 = a 1 + (3 – 1)d 3 – 1 = 2
sides.
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= a +- (n
– 1)d n
=3
n
Page 31
1
Substitution
a 3 = a 1 + 2d Associative Property of Addition
a 3 = a 1 + (3 – 1)d 3 – 1 = 2
Simplify.
a n = a 1 + (n Sequences
– 1)d n =and
3 Series
10-2 Arithmetic
81. PROOF Derive a sum formula that does not
include a 1.
82. PROOF Derive the Alternate Sum Formula using
the General Sum Formula.
SOLUTION: SOLUTION: General sum formula
General sum formula
a n = a 1 + (n – 1)d Formula for nth term
a n = a 1 + (n – 1)d Formula for nth term
a n – (n – 1)d = a 1 Subtract (n – 1)d from both
sides.
Substitution
Simplify.
Substitution
Simplify.
ANSWER: General sum formula
ANSWER: General sum formula
a n = a 1 + (n – 1)d Formula for nth term
Substitution
a n = a 1 + (n – 1)d Formula for nth term
Simplify.
a n – (n – 1)d = a 1 Subtract (n – 1)d from both
sides
.
Substitution
83. SAT/ACT The measures of the angles of a
triangle form an arithmetic sequence. If the
measure of the smallest angle is 36°, what is the measure of the largest angle?
Simplify.
82. PROOF Derive the Alternate Sum Formula using
the General Sum Formula.
A 54°
B 75°
SOLUTION: C 84°
General sum formula
D 90°
a n = a 1 + (n – 1)d Formula for nth term
E 97°
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Page 32
Substitution
SOLUTION: The sum of the interior angle of a triangle is 180°.
Since the measures of the angles of a triangle form
Option C is the correct answer.
Substitution
ANSWER: C
10-2 Arithmetic Sequences and
Series
Simplify.
83. SAT/ACT The measures of the angles of a
triangle form an arithmetic sequence. If the
measure of the smallest angle is 36°, what is the measure of the largest angle?
84. The area of a triangle is
and the height is q
+ 4. Which expression best describes the triangle’s
length?
F (q + 1)
G (q + 2)
H (q – 3)
A 54°
J (q – 4)
B 75°
C 84°
SOLUTION: D 90°
The area of the triangle is
.
E 97°
The length of the triangle is
SOLUTION: The sum of the interior angle of a triangle is 180°.
Since the measures of the angles of a triangle form
an arithmetic sequence and the smallest angle is
36°, the other two angles are 36° + d and 36° + 2d.
36 + 36 + d + 36 + 2d = 180
d = 24
Therefore, the largest angle is 36° + 48° = 84°.
Option C is the correct answer.
Option J is the correct answer.
ANSWER: J
85. The expression
is equivalent to
ANSWER: C
A
84. The area of a triangle is
and the height is q
+ 4. Which expression best describes the triangle’s
length?
B
C
F (q + 1)
G (q + 2)
D
H (q – 3)
J (q – 4)
SOLUTION: SOLUTION: eSolutions Manual - Powered by Cognero
The area of the triangle is
Page 33
.
Option A is the correct answer.
Option A is the correct answer.
ANSWER: J
10-2 Arithmetic
Sequences and Series
85. The expression
is equivalent to
ANSWER: A
86. SHORT RESPONSE Trevor can type a 200-word
essay in 6 hours. Minya can type the same essay in
hours. If they work together, how many hours
A
will it take them to type the essay?
B
SOLUTION: Trevor can type
C
Minya can type
words in an hour.
words in an hour.
D
They type
words together in an
hour.
SOLUTION: Option A is the correct answer.
ANSWER: A
86. SHORT RESPONSE Trevor can type a 200-word
essay in 6 hours. Minya can type the same essay in
They will type the essay in
hours if they work together.
hours. If they work together, how many hours
will it take them to type the essay?
ANSWER: SOLUTION: Trevor can type
words in an hour.
Determine whether each sequence is
arithmetic. Write yes or no.
Minya can type
words in an hour.
They type
hour.
87. –6, 4, 14, 24, …
words together in an
SOLUTION: Since there is a common difference between the
consecutive terms, this is an arithmetic sequence.
ANSWER: yes
eSolutions Manual - Powered by Cognero
Page 34
ANSWER: ANSWER: no
10-2 Arithmetic Sequences and Series
Determine whether each sequence is
arithmetic. Write yes or no.
Solve each system of inequalities by graphing.
87. –6, 4, 14, 24, …
90. SOLUTION: Since there is a common difference between the
consecutive terms, this is an arithmetic sequence.
SOLUTION: ANSWER: yes
88. ANSWER: SOLUTION: Since there is a common difference between the
consecutive terms, this is an arithmetic sequence.
ANSWER: yes
89. 10, 8, 5, 1, ...
SOLUTION: Since there is no common difference between the
consecutive terms, this is not an arithmetic
sequence.
ANSWER: no
91. SOLUTION: Solve each system of inequalities by graphing.
90. SOLUTION: ANSWER: eSolutions Manual - Powered by Cognero
Page 35
10-2 Arithmetic Sequences and Series
91. 92. SOLUTION: SOLUTION: ANSWER: ANSWER: 92. SOLUTION: 93. PHYSICS The distance a spring stretches is
related to the mass attached to the spring. This is
represented by d = km, where d is the distance, m is
the mass, and k is the spring constant. When two
springs with spring constants k 1 and k 2 are attached
in a series, the resulting spring constant k is found
by the equation
a. If one spring with constant of 12 centimeters per
gram is attached in a series with another spring with
constant of 8 centimeters per gram, find the
resultant spring constant.
ANSWER: b. If a 5-gram object is hung from the series of
springs, how far will the springs stretch? Is this
answer reasonable in this context?
eSolutions Manual - Powered by Cognero
Page 36
resultant spring constant.
would stretch the first spring 60 cm and would
stretch the second spring 40 cm. The object would
have to stretch the combined springs less than it
would stretch either of the springs individually.
b. If a 5-gram object is hung from the series of
springs, howSequences
far will the springs
stretch? Is this
10-2 Arithmetic
and Series
answer reasonable in this context?
Graph each function. State the domain and
range.
94. SOLUTION: Graph the function.
SOLUTION: a. Given k 1 = 12 and k 2 = 8.
Substitute the values and evaluate the value of k.
The function is defined for all values of x.
Therefore, the domain is D = {all real numbers}.
The value of the f (x) tends to 0 as x tends to –∞.
The value of the f (x) tends to ∞ as x tends to ∞.
b. Substitute 4.8 and 5 for k and m respectively in
the equation d = km.
Therefore, the range of the function is R ={f (x) | f
(x) > 0}.
ANSWER: The answer is reasonable. The object would stretch
the first spring 60 cm and would stretch the second
spring 40 cm. The object would have to stretch the
combined springs less than it would stretch either of
the springs individually.
ANSWER: a. 4.8 cm/g
D = {all real numbers}, R = {f (x) | f (x) > 0}
b. 24 cm; The answer is reasonable. The object
would stretch the first spring 60 cm and would
stretch the second spring 40 cm. The object would
have to stretch the combined springs less than it
would stretch either of the springs individually.
Graph each function. State the domain and
range.
eSolutions Manual - Powered by Cognero
95. SOLUTION: Graph the function.
Page 37
D = {all real numbers}, R = {f (x) | f (x) > 0}
10-2 Arithmetic Sequences and Series
D = {all real numbers}, R = {f (x) | f (x) > 3}
95. SOLUTION: Graph the function.
96. SOLUTION: Graph the function.
The function is defined for all values of x.
The function is defined for all values of x.
Therefore, the domain is D = {all real numbers}.
Therefore, the domain is D = {all real numbers}.
The value of the f (x) tends to 3 as x tends to –∞.
The value of the f (x) tends to ∞ as x tends to –∞.
The value of the f (x) tends to ∞ as x tends to ∞.
The value of the f (x) tends to –1 as x tends to ∞.
Therefore, the range of the function is R ={f (x) | f
(x) > 3}.
ANSWER: Therefore, the range of the function is R ={f (x) | f
(x) > –1}.
ANSWER: D = {all real numbers}, R = {f (x) | f (x) > 3}
D = {all real numbers}, R = {f (x) | f (x) > –1}
Solve each equation. Round to the nearest tenthousandth.
96. SOLUTION: Graph the function.
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x
97. 5 = 52
SOLUTION: Page 38
10-2 Arithmetic
Sequences and Series
D = {all real numbers}, R = {f (x) | f (x) > –1}
Solve each equation. Round to the nearest tenthousandth.
ANSWER: 0.5537
n+2
99. 3
= 14.5
SOLUTION: x
97. 5 = 52
SOLUTION: ANSWER: 0.4341
ANSWER: 2.4550
d –4
100. 16
3–d
=3
3p
98. 4
SOLUTION: = 10
SOLUTION: ANSWER: 0.5537
n+2
99. 3
= 14.5
ANSWER: 3.7162
SOLUTION: ANSWER: 0.4341
eSolutions Manual - Powered by Cognero
d –4
100. 16
3–d
=3
Page 39
10-3 Geometric Sequences and Series
ANSWER: 2046
1. CCSS REGULARITY Dean is making a family
tree for his grandfather. He was able to trace many
generations. If Dean could trace his family back 10
generations, starting with his parents how many
ancestors would there be?
SOLUTION: Let a 1 = 2, r = 2, and n = 10.
Write an equation for the nth term of each
geometric sequence.
2. 2, 4, 8, …
SOLUTION: ANSWER: 3. 18, 6, 2, …
SOLUTION: Dean will identify 2046 ancestors..
ANSWER: 2046
Write an equation for the nth term of each
geometric sequence.
ANSWER: 2. 2, 4, 8, …
SOLUTION: 4. –4, 16, –64, …
SOLUTION: ANSWER: eSolutions Manual - Powered by Cognero
3. 18, 6, 2, …
Page 1
ANSWER: ANSWER: ANSWER: 10-3 Geometric Sequences and Series
4. –4, 16, –64, …
SOLUTION: 6. SOLUTION: Find a 1.
ANSWER: 5. a 2 = 4, r = 3
SOLUTION: Find a 1.
Write the equation.
ANSWER: Write the equation.
7. a 2 = –96, r = –8
ANSWER: SOLUTION: Find a 1.
6. Write the equation.
SOLUTION: Find a 1.
eSolutions Manual - Powered by Cognero
ANSWER: Page 2
ANSWER: ANSWER: 1, 4, 16 or –1, 4, –16]
10-3 Geometric Sequences and Series
7. a 2 = –96, r = –8
9. SOLUTION: Find a 1.
SOLUTION: Given a 1 = 0.20 and a 5 = 125.
The geometric means are 1, 5, 25 or –1,5,–25.
Write the equation.
ANSWER: 1, 5, 25 or –1, 5, –25
ANSWER: Find the geometric means of each sequence.
10. GAMES Miranda arranges some rows of dominoes
so that after she knocks over the first one, each
domino knocks over two more dominoes when it
falls. If there are ten rows, how many dominoes does
Miranda use?
8. SOLUTION: SOLUTION: Here a 1 = 0.25 and a 5 = 64.
ANSWER: 1023
Find the sum of each geometric series.
The geometric means are 1, 4, 16 or –1, 4, –16.
ANSWER: 1, 4, 16 or –1, 4, –16]
11. SOLUTION: Given r = 4.
There are 6 – 1 + 1 or 6 terms, so n = 6.
1–1
a 1 = 3(4)
9. SOLUTION: Given a 1 = 0.20 and a 5 = 125.
eSolutions Manual - Powered by Cognero
=3
Find the sum.
Page 3
ANSWER: 4095
ANSWER: 10-31023
Geometric Sequences and Series
Find the sum of each geometric series.
12. 11. SOLUTION: Given r = 4.
SOLUTION: Given r = 4.
There are 8 – 1 + 1 or 8 terms, so n = 8.
There are 6 – 1 + 1 or 6 terms, so n = 6.
1–1
a 1 = 3(4)
=3
Find the sum.
Find the sum.
ANSWER: 4095
ANSWER: 7.96875
12. Find a 1 for each geometric series described.
SOLUTION: Given r = 4.
There are 8 – 1 + 1 or 8 terms, so n = 8.
13. SOLUTION: Find the sum.
Substitute the corresponding values and solve for a 1.
eSolutions Manual - Powered by Cognero
Page 4
ANSWER: ANSWER: 10-37.96875
Geometric Sequences and Series
Find a 1 for each geometric series described.
14. 13. SOLUTION: SOLUTION: Substitute the corresponding values and solve for a 1.
Substitute the corresponding values and solve for a 1.
ANSWER: ANSWER: 15. 14. SOLUTION: SOLUTION: Substitute the corresponding values and solve for a 1.
Substitute the corresponding values and solve for a 1.
eSolutions Manual - Powered by Cognero
Page 5
ANSWER: 512
ANSWER: ANSWER: 512
10-3 Geometric Sequences and Series
16. 15. SOLUTION: SOLUTION: Substitute the corresponding values and solve for a 1.
Substitute the corresponding values and solve for a 1.
ANSWER: 512
16. SOLUTION: ANSWER: 81
Substitute the corresponding values and solve for a 1.
17. WEATHER Heavy rain in Brieanne’s town caused
the river to rise. The river rose three inches the first
day, and each day after rose twice as much as the
previous day. How much did the river rise in five
days?
SOLUTION: Let a 1 = 3, r = 2, and n =5.
eSolutions Manual - Powered by Cognero
Page 6
ANSWER: 10-381
Geometric Sequences and Series
ANSWER: 93 in.
17. WEATHER Heavy rain in Brieanne’s town caused
the river to rise. The river rose three inches the first
day, and each day after rose twice as much as the
previous day. How much did the river rise in five
days?
Find a n for each geometric sequence.
18. SOLUTION: SOLUTION: Let a 1 = 3, r = 2, and n =5.
The nth term of a geometric sequence is
.
ANSWER: or 0.5859375
The river rose 93 inches in 5 days.
19. ANSWER: 93 in.
SOLUTION: The nth term of a geometric sequence is
.
Find a n for each geometric sequence.
18. SOLUTION: The nth term of a geometric sequence is
ANSWER: 25
.
20. SOLUTION: ANSWER: eSolutions Manual
- Powered by Cognero
or 0.5859375
The nth term of a geometric sequence is
.
Page 7
ANSWER: 512
ANSWER: 10-325
Geometric Sequences and Series
22. BIOLOGY A certain bacteria grows at a rate of 3
cells every 2 minutes. If there were 260 cells initially,
how many are there after 21 minutes?
20. SOLUTION: The nth term of a geometric sequence is
.
SOLUTION: Given, a 1 = 260, n = 21 and r =
.
ANSWER: 162
ANSWER: 864,567
21. Write an equation for the nth term of each
geometric sequence.
SOLUTION: The nth term of a geometric sequence is
.
23. –3, 6, –12, …
SOLUTION: ANSWER: 512
22. BIOLOGY A certain bacteria grows at a rate of 3
cells every 2 minutes. If there were 260 cells initially,
how many are there after 21 minutes?
ANSWER: n –1
a n = (–3)(–2)
SOLUTION: Given, a 1 = 260, n = 21 and r =
.
24. 288, –96, 32, …
SOLUTION: ANSWER: 864,567
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Manual - Powered by Cognero
Page 8
Write an equation for the nth term of each
ANSWER: ANSWER: n –1
a n = (–3)(–2)
10-3 Geometric Sequences and Series
24. 288, –96, 32, …
n –1
a n = (–1)(–1)
26. SOLUTION: SOLUTION: ANSWER: ANSWER: 25. –1, 1, –1, …
SOLUTION: 27. SOLUTION: ANSWER: n –1
a n = (–1)(–1)
26. ANSWER: SOLUTION: 28. eSolutions
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ANSWER: SOLUTION: Page 9
ANSWER: ANSWER: 10-3 Geometric Sequences and Series
30. a 4 = –8, r = 0.5
28. SOLUTION: SOLUTION: Find a 1.
ANSWER: 29. a 3 = 28, r = 2
SOLUTION: Find a 1.
Write the equation.
ANSWER: a n = –64(0.5)
n –1
31. a 6 = 0.5, r = 6
SOLUTION: Find a 1.
Write the equation.
ANSWER: Write the equation.
30. a 4 = –8, r = 0.5
SOLUTION: eSolutions Manual - Powered by Cognero
Find a 1.
ANSWER: Page 10
ANSWER: ANSWER: n –1
a n = –64(0.5)
10-3 Geometric
Sequences and Series
31. a 6 = 0.5, r = 6
32. SOLUTION: Find a 1.
SOLUTION: Find a 1.
Write the equation.
Write the equation.
ANSWER: ANSWER: 32. SOLUTION: Find a 1.
33. SOLUTION: Find a 1.
Write the equation.
Write the equation
.
eSolutions Manual - Powered by Cognero
Page 11
ANSWER: ANSWER: 10-3 Geometric Sequences and Series
34. a 4 = 80, r = 4
33. SOLUTION: Find a 1.
SOLUTION: Find a 1.
Write the equation.
Write the equation
.
ANSWER: ANSWER: Find the geometric means of each sequence.
34. a 4 = 80, r = 4
35. SOLUTION: Find a 1.
SOLUTION: Since there are three terms between the first and last
term, there are 3 + 2 or 5 total terms, so n = 5.
Given a 1 = 810 and a 5 = 10.
Write the equation.
The geometric means are 270, 90, 30 or –270, 90, –
30.
eSolutions Manual - Powered by Cognero
ANSWER: 270, 90, 30 or –270, 90, –30
Page 12
ANSWER: ANSWER: 270, 90, 30 or –270, 90, –30
10-3 Geometric Sequences and Series
Find the geometric means of each sequence.
36. 35. SOLUTION: Since there are three terms between the first and last
term, there are 3 + 2 or 5 total terms, so n = 5.
Given a 1 = 640 and a 5 = 2.5.
SOLUTION: Since there are three terms between the first and last
term, there are 3 + 2 or 5 total terms, so n = 5.
Given a 1 = 810 and a 5 = 10.
The geometric means are 160, 40, 10 or –160, 40, –
10.
The geometric means are 270, 90, 30 or –270, 90, –
30.
ANSWER: 160, 40, 10 or –160, 40, –10
ANSWER: 270, 90, 30 or –270, 90, –30
37. 36. SOLUTION: Since there are three terms between the first and last
term, there are 3 + 2 or 5 total terms, so n = 5.
Given a 1 = 640 and a 5 = 2.5.
SOLUTION: Since there are three terms between the first and last
term, there are 3 + 2 or 5 total terms, so n = 5.
Given a 1 =
and a 5 =
.
The geometric means are 160, 40, 10 or –160, 40, –
10.
ANSWER: 160, 40, 10 or –160, 40, –10
eSolutions Manual - Powered by Cognero
37. The geometric means are
or .
ANSWER: or
Page 13
ANSWER: ANSWER: 40, 10 orSequences
–160, 40, –10
10-3160,
Geometric
and Series
or
37. 38. SOLUTION: Since there are three terms between the first and last
term, there are 3 + 2 or 5 total terms, so n = 5.
SOLUTION: Since there are three terms between the first and last
term, there are 3 + 2 or 5 total terms, so n = 5.
Given a 1 =
and a 5 =
and a 5 =
Given a 1 =
.
or The geometric means are
.
or The geometric means are
.
.
ANSWER: ANSWER: or
or
39. Find two geometric means between 3 and 375.
38. SOLUTION: Since there are three terms between the first and last
term, there are 3 + 2 or 5 total terms, so n = 5.
Given a 1 = 3 and a 4 = 375.
Given a 1 =
SOLUTION: Since there are two terms between the first and last
term, there are 2 + 2 or 4 total terms, so n = 4.
and a 5 =
.
The geometric means are
.
ANSWER: 15, 75
eSolutions Manual - Powered by Cognero
The geometric means are
or Page 14
The geometric means are
ANSWER: .
ANSWER: –8, 4
or
10-3 Geometric Sequences and Series
39. Find two geometric means between 3 and 375.
SOLUTION: Since there are two terms between the first and last
term, there are 2 + 2 or 4 total terms, so n = 4.
41. CCSS PERSEVERANCE A certain water
filtration system can remove 70% of the
contaminants each time a sample of water is passed
through it. If the same water is passed through the
system four times, what percent of the original
contaminants will be removed from the water
sample?
Given a 1 = 3 and a 4 = 375.
SOLUTION: Let 100% contaminated water passes for the
purification.
That is, a 0 = 1.
After the filtration, the system removes 70% of the
contaminant.
That is 30% of contamination still in the water.
Therefore, r = 100% – 70% = 30% or 0.3.
The number of times (n) is 4.
The geometric means are
.
ANSWER: 15, 75
40. Find two geometric means between 16 and –2.
SOLUTION: Since there are two terms between the first and last
term, there are 2 + 2 or 4 total terms, so n = 4.
Therefore, after four times of filtration, 1 – 0.081 =
0.9919 or 99.19% of the original contaminants will be
removed.
Given a 1 = 16 and a 4 = –2.
ANSWER: 99.19%
Find the sum of each geometric series.
42. The geometric means are
.
SOLUTION: Find the sum.
ANSWER: –8, 4
41. CCSS PERSEVERANCE A certain water
filtration system can remove 70% of the
contaminants each time a sample of water is passed
through it. If the same water is passed through the
system four times, what percent of the original
contaminants
will be
eSolutions
Manual - Powered
byremoved
Cognero from the water
sample?
ANSWER: 53.9918
Page 15
ANSWER: 10-399.19%
Geometric Sequences and Series
ANSWER: 31.9375
Find the sum of each geometric series.
44. 42. SOLUTION: Find the sum.
SOLUTION: Find the sum.
ANSWER: 831.855
ANSWER: 53.9918
45. 43. SOLUTION: Find the sum.
SOLUTION: Find the sum.
ANSWER: 31.9375
ANSWER: 9707.82
44. SOLUTION: Find the sum.
46. VACUUMS A vacuum claims to pick up 80% of the
dirt every time it is run over the carpet. Assuming
this is true, what percent of the original amount of
dirt is picked up after the seventh time the vacuum is
run over the carpet?
SOLUTION: Let 100% of dirt be in the carpet.
eSolutions Manual - Powered by Cognero
That is, a 0 = 1.
Page 16
A vacuum claims to pick up 80% of the dirt every
ANSWER: 10-39707.82
Geometric Sequences and Series
ANSWER: 99.99%
46. VACUUMS A vacuum claims to pick up 80% of the
dirt every time it is run over the carpet. Assuming
this is true, what percent of the original amount of
dirt is picked up after the seventh time the vacuum is
run over the carpet?
Find the sum of each geometric series.
47. SOLUTION: Let 100% of dirt be in the carpet.
SOLUTION: Given r = –3.
There are 7 – 1 + 1 or 7 terms, so n = 7.
1–1
a 1 = 4(–3)
=4
That is, a 0 = 1.
Find the sum.
A vacuum claims to pick up 80% of the dirt every
time it is run over the carpet.
That is 20% of dirt still in the carpet.
Therefore, r = 100% – 80% = 20% or 0.2.
The number of times (n) is 7.
Therefore, after seven times of cleaning, 1 –
0.0000128 = 0.9999872 or 99.99872% of the original
dirt is picked up.
ANSWER: 2188
ANSWER: 99.99%
Find the sum of each geometric series.
48. SOLUTION: Given r = –2. 47. There are 8 – 1 + 1 or 8 terms, so n = 8.
SOLUTION: Given r = –3.
There are 7 – 1 + 1 or 7 terms, so n = 7.
1–1
a 1 = 4(–3)
=4
a 1 = (–3)(–2)
1–1
= –3
Find the sum.
Find the sum.
eSolutions Manual - Powered by Cognero
Page 17
ANSWER: 2188
10-3 Geometric Sequences and Series
48. ANSWER: 255
49. SOLUTION: Given r = –2. SOLUTION: Given r = 4.
There are 8 – 1 + 1 or 8 terms, so n = 8.
There are 9 – 1 + 1 or 9 terms, so n = 9.
1–1
a 1 = (–3)(–2)
= –3
1–1
a 1 = (–1)(4)
Find the sum.
Find the sum.
= –1
ANSWER: 255
ANSWER: –87, 381
49. 50. SOLUTION: Given r = 4.
SOLUTION: Given r = –1.
There are 9 – 1 + 1 or 9 terms, so n = 9.
There are 10 – 1 + 1 or 10 terms, so n = 10.
1–1
a 1 = (–1)(4)
= –1
1–1
a 1 = 5(–1)
Find the sum.
Find the sum.
eSolutions Manual - Powered by Cognero
=5
Page 18
ANSWER: 0
ANSWER: 381
10-3–87,
Geometric
Sequences and Series
Find a 1 for each geometric series described.
50. 51. S n = –2912, r = 3, n = 6
SOLUTION: Given r = –1.
SOLUTION: There are 10 – 1 + 1 or 10 terms, so n = 10.
a 1 = 5(–1)
1–1
=5
Find the sum.
Substitute the corresponding values and solve for a 1.
ANSWER: –8
ANSWER: 0
52. S n = –10,922, r = 4, n = 7
Find a 1 for each geometric series described.
SOLUTION: 51. S n = –2912, r = 3, n = 6
SOLUTION: Substitute the corresponding values and solve for a 1.
Substitute the corresponding values and solve for a 1.
ANSWER: –2
eSolutions
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ANSWER: Page 19
53. ANSWER: –8
10-3 Geometric Sequences and Series
52. S n = –10,922, r = 4, n = 7
SOLUTION: ANSWER: 64
54. SOLUTION: Substitute the corresponding values and solve for a 1.
Substitute the corresponding values and solve for a 1.
ANSWER: –2
ANSWER: 1458
53. SOLUTION: 55. a n = 1024, r = 8, n = 5
SOLUTION: Substitute the corresponding values and solve for a 1.
The nth term of a geometric sequence is
.
ANSWER: 0.25
56. a n = 1875, r = 5, n = 7
ANSWER: 64
SOLUTION: The nth term of a geometric sequence is
.
54. eSolutions Manual - Powered by Cognero
SOLUTION: Page 20
ANSWER: 10-30.25
Geometric Sequences and Series
ANSWER: 193.75 ft
56. a n = 1875, r = 5, n = 7
SOLUTION: The nth term of a geometric sequence is
.
58. CHEMISTRY Radon has a half-life of about 4
days. This means that about every 4 days, half of the
mass of radon decays into another element. How
many grams of radon remain from an initial 60 grams
after 4 weeks?
SOLUTION: Given a 0 = 60 g,
,n=7
ANSWER: After 4 weeks the amount of radon remain is about
0.46875 g.
57. SCIENCE One minute after it is released, a gas-
filled balloon has risen 100 feet. In each succeeding
minute, the balloon rises only 50% as far as it rose
in the previous minute. How far will it rise in 5
minutes?
SOLUTION: Let a 1 = 100, r = 50% or 0.5, and n = 5.
ANSWER: about 0.46875 g
59. CCSS REASONING A virus goes through a
computer, infecting the files. If one file was infected
initially and the total number of files infected doubles
every minute, how many files will be infected in 20
minutes?
SOLUTION: Given, a 1 = 1, r = 200% or 2, n = 20.
ANSWER: 524,288
The balloon will rise 193.75 feet.
ANSWER: 193.75 ft
60. GEOMETRY In the figure, the sides of each
equilateral triangle are twice the size of the sides of
its inscribed triangle. If the pattern continues, find the
sum of the perimeters of the first eight triangles.
58. CHEMISTRY Radon has a half-life of about 4
4 days, half of the
mass of radon decays into another element. How
many grams of radon remain from an initial 60 grams
eSolutions
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days.Manual
This -means
that
about every
Page 21
ANSWER: 10-3524,288
Geometric Sequences and Series
ANSWER: about 119.5 cm
60. GEOMETRY In the figure, the sides of each
equilateral triangle are twice the size of the sides of
its inscribed triangle. If the pattern continues, find the
sum of the perimeters of the first eight triangles.
61. PENDULUMS The first swing of a pendulum
travels 30 centimeters. If each subsequent swing
travels 95% as far as the previous swing, find the
total distance traveled by the pendulum after the 30th
swing.
SOLUTION: Given a 1 = 30 cm, r = 95 %, n = 30.
Find the sum.
SOLUTION: Here a 1 = 60 cm, r = 0.5, n = 8.
The distance traveled is about 471 cm.
The sum of the perimeters of the first eight triangles
is about 119.5 cm.
ANSWER: about 119.5 cm
61. PENDULUMS The first swing of a pendulum
travels 30 centimeters. If each subsequent swing
travels 95% as far as the previous swing, find the
total distance traveled by the pendulum after the 30th
swing.
ANSWER: about 471 cm
62. PHONE CHAINS A school established a phone
chain in which every staff member calls two other
staff members to notify them when the school closes
due to weather. The first round of calls begins with
the superintendent calling both principals. If there are
94 total staff members and employees at the school,
how many rounds of calls are there?
SOLUTION: Given, a 1 = 2, r = 2 and a n = 94
SOLUTION: Given a 1 = 30 cm, r = 95 %, n = 30.
Find n.
Find the sum.
There are 7 rounds of calls.
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distance
traveled
is about
471 cm.
ANSWER: 7
Page 22
ANSWER: 471 cmSequences and Series
10-3about
Geometric
SOLUTION: Given
a. Substitute 5, 1.1 and 10 for a 1, r and n
respectively in
62. PHONE CHAINS A school established a phone
chain in which every staff member calls two other
staff members to notify them when the school closes
due to weather. The first round of calls begins with
the superintendent calling both principals. If there are
94 total staff members and employees at the school,
how many rounds of calls are there?
then simplify.
Substitute 5, 1.1 and 20 for a 1, r and n respectively in
then simplify.
SOLUTION: Given, a 1 = 2, r = 2 and a n = 94
Substitute 5, 1.1 and 40 for a 1, r and n respectively in
Find n.
then simplify.
b. Substitute 5, 1.1 and 40 for a 1, r and n
respectively in
then simplify.
There are 7 rounds of calls.
ANSWER: 7
63. TELEVISIONS High Tech Electronics advertises a
weekly installment plan for the purchase of a popular
brand of high definition television. The buyer pays $5
at the end of the first week, $5.50 at the end of the
second week, $6.05 at the end of the third week, and
so on got one year. Assume 1 year = 52 weeks)
c. Each payment made is rounded to the nearest
penny, so the sum of the payments will actually be
more than the sum found in part b.
ANSWER: a. $11.79, $30.58, $205.72
b. $7052.15
a. What will the payments be at the end of the 10th,
20th, and 40th weeks?
c. Each payment made is rounded to the nearest
penny, so the sum of the payments will actually be
more than the sum found in part b.
b. Find the total cost of the TV.
c. Why is the cost found in part b not entirely
accurate?
64. PROOF Derive the General Sum Formula using
theAlternate Sum Formula.
SOLUTION: Given
a. Substitute 5, 1.1 and 10 for a 1, r and n
SOLUTION: respectively in
then simplify.
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Page 23
c. Each payment made is rounded to the nearest
penny, so the sum of the payments will actually be
than theSequences
sum found in
partSeries
b.
10-3more
Geometric
and
64. PROOF Derive the General Sum Formula using
theAlternate Sum Formula.
SOLUTION: 65. PROOF Derive a sum formula that does not include
a 1.
SOLUTION: ANSWER: 65. PROOF Derive a sum formula that does not include
a 1.
ANSWER: SOLUTION: 66. OPEN ENDED Write a geometric series for which
and n = 6.
ANSWER: SOLUTION: Sample answer:
eSolutions Manual - Powered by Cognero
Page 24
ANSWER: terms for both series will be identical (a 1 in the first
series will equal a 0 in the second series, and so on),
and the series will be equal to each other.
10-3 Geometric Sequences and Series
66. OPEN ENDED Write a geometric series for which
and n = 6.
68. PROOF Prove the formula for the nth term of a
geometric sequence.
SOLUTION: Sample answer:
SOLUTION: Sample answer: Let a n = the nth term of the
sequence and r = the common ratio.
ANSWER: Sample answer:
67. REASONING Explain how
needs to be altered to refer to the same series if k = 1 changes to
k = 0. Explain your reasoning.
ANSWER: Sample answer: Let a n = the nth term of the
sequence and r = the common ratio.
SOLUTION: Sample answer: n – 1 needs to change to n, and the
10 needs to change to a 9. When this happens, the
terms for both series will be identical (a 1 in the first
series will equal a 0 in the second series, and so on),
and the series will be equal to each other.
ANSWER: Sample answer: n – 1 needs to change to n, and the
10 needs to change to a 9. When this happens, the
terms for both series will be identical (a 1 in the first
69. CHALLENGE The fifth term of a geometric
series will equal a 0 in the second series, and so on),
sequence is
and the series will be equal to each other.
term is 702, what is the eighth term?
68. PROOF Prove the formula for the nth term of a
geometric sequence.
SOLUTION: Sample answer: Let a n = the nth term of the
sequence and r = the common ratio.
th of the eighth term. If the ninth
SOLUTION: Given a 5 =
and a 9 = 702.
Find the value of r.
eSolutions Manual - Powered by Cognero
Page 25
ANSWER: 234
10-3 Geometric Sequences and Series
69. CHALLENGE The fifth term of a geometric
70. CHALLENGE Use the fact that h is the geometric
4
sequence is
th of the eighth term. If the ninth
term is 702, what is the eighth term?
mean between x and y in the figure to find h in
terms of x and y.
SOLUTION: Given a 5 =
and a 9 = 702.
Find the value of r.
SOLUTION: The triangle ADC and the triangle CDB are similar.
Therefore,
and .
Consider the triangle ABC.
Find the value of a 8.
ANSWER: 234
ANSWER: 2 2
xy
70. CHALLENGE Use the fact that h is the geometric
4
mean between x and y in the figure to find h in
terms of x and y.
71. OPEN ENDED Write a geometric series with 6
terms and a sum of 252.
SOLUTION: Sample answer: 4 + 8 + 16 + 32 + 64 + 128
SOLUTION: The triangle ADC and the triangle CDB are similar.
Therefore,
and .
ANSWER: Sample answer: 4 + 8 + 16 + 32 + 64 + 128
72. WRITING IN MATH How can you classify a
sequence? Explain your reasoning.
Consider the triangle ABC.
SOLUTION: eSolutions Manual - Powered by Cognero
Sample answer: A series is arithmetic if every pair
Page 26
of consecutive terms shares a common difference.
A series is geometric if every pair of consecutive
terms shares a common ratio. If the series displays
ANSWER: answer:
4 + 8 + 16and
+ 32Series
+ 64 + 128
10-3Sample
Geometric
Sequences
72. WRITING IN MATH How can you classify a
sequence? Explain your reasoning.
SOLUTION: Sample answer: A series is arithmetic if every pair
of consecutive terms shares a common difference.
A series is geometric if every pair of consecutive
terms shares a common ratio. If the series displays
both qualities, then it is both arithmetic and
geometric. If the series displays neither quality, then
it is neither geometric nor arithmetic.
ANSWER: B
74. The first term of a geometric series is 5, and the
common ratio is −2. How many terms are in the
series if its sum is
F 5
G 9
H 10
J 12
SOLUTION: ANSWER: Sample answer: A series is arithmetic if every pair
of consecutive terms shares a common difference.
A series is geometric if every pair of consecutive
terms shares a common ratio. If the series displays
both qualities, then it is both arithmetic and
geometric. If the series displays neither quality, then
it is neither geometric nor arithmetic.
73. Which of the following is closest to
A 1.8
Therefore, J is the correct answer.
B 1.9
C 2.0
D 2.1
SOLUTION: ANSWER: J 75. SHORT RESPONSE Danette has a savings
account. She withdraws half of the contents every
year. After 4 years, she has $2000 left. How much
did she have in the savings account originally?
Option B is the correct answer.
ANSWER: B
SOLUTION: Given n = 4,a 4 = 2000 r = 0.5
74. The first term of a geometric series is 5, and the
common ratio is −2. How many terms are in the
series if its sum is
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eSolutions
G 9
Page 27
ANSWER: 10-3J Geometric Sequences and Series
75. SHORT RESPONSE Danette has a savings
account. She withdraws half of the contents every
year. After 4 years, she has $2000 left. How much
did she have in the savings account originally?
ANSWER: $32,000
76. SAT/ACT The curve below could be part of the
graph of which function?
SOLUTION: Given n = 4,a 4 = 2000 r = 0.5
A
B
C
D
She have invested (a 0) $32,000.
ANSWER: $32,000
76. SAT/ACT The curve below could be part of the
graph of which function?
E
SOLUTION: For x = 0, the value of y become 0 in the equation
. Therefore, this is not a correct answer.
The Graph of the equations
represents are a parabola and a linear equation. So they are not the correct answer.
The graph of
increases as x increases;
the given graph is decreasing.
Therefore, option E is the correct answer.
ANSWER: E
B
A
C
D
E
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SOLUTION: For x = 0, the value of y become 0 in the equation
77. MONEY Elena bought a high-definition LCD
television at the electronics store. She paid $200
immediately and $75 each month for a year and a
half. How much did Elena pay in total for the TV?
SOLUTION: She paid $75 for 18 months.
So, she paid in total of 200 + (75 × 18) or $1550 for Page 28
TV.
ANSWER: Neither; there is no common difference or ratio.
ANSWER: 10-3EGeometric Sequences and Series
77. MONEY Elena bought a high-definition LCD
television at the electronics store. She paid $200
immediately and $75 each month for a year and a
half. How much did Elena pay in total for the TV?
79. SOLUTION: Find any common difference or ratio in the sequence.
SOLUTION: She paid $75 for 18 months.
So, she paid in total of 200 + (75 × 18) or $1550 for TV.
ANSWER: $1550
Determine whether each sequence is
arithmetic, geometric, or neither. Explain your
reasoning.
There common difference is
.
Therefore, this is an arithmetic sequence.
78. ANSWER: Arithmetic; the common difference is
SOLUTION: Find any common difference or ratio in the sequence.
.
80. SOLUTION: Find any common difference or ratio in the sequence.
There is no common difference or ratio.
ANSWER: Neither; there is no common difference or ratio.
There is no common difference or ratio.
79. SOLUTION: Find any common difference or ratio in the sequence.
ANSWER: Neither; there is no common difference or ratio.
Find the center and radius of each circle. Then
graph the circle.
2
2
81. (x – 3) + (y – 1) = 25
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SOLUTION: Page 29
ANSWER: thereSequences
is no common
10-3Neither;
Geometric
anddifference
Series or ratio.
Find the center and radius of each circle. Then
graph the circle.
2
2
81. (x – 3) + (y – 1) = 25
SOLUTION: Rewrite the equation in the standard form of the
circle.
2
2
82. (x + 3) + (y + 7) = 81
SOLUTION: Rewrite the equation in the standard form of the
circle.
The center and the radius of the circle are (–3, –7)
and 9 units respectively.
Graph the equation.
The center and the radius of the circle are (3, 1) and
5 units respectively.
Graph the equation.
ANSWER: (–3, –7), 9 units
ANSWER: (3, 1), 5 units
2
2
83. (x – 3) + (y + 7) = 50
2
2
82. (x + 3) + (y + 7) = 81
SOLUTION: Rewrite the equation in the standard form of the
circle.
SOLUTION: Rewrite the equation in the standard form of the
circle.
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Page 30
The center and the radius of the circle are (3, –7)
and
units respectively.
10-3 Geometric Sequences and Series
2
2
83. (x – 3) + (y + 7) = 50
84. Suppose y varies jointly as x and z. Find y when x =
9 and z = –5, if y = –90 when z = 15 and x = –6.
SOLUTION: Rewrite the equation in the standard form of the
circle.
SOLUTION: Given
.
Find the value of k.
The center and the radius of the circle are (3, –7)
and
units respectively.
Graph the equation.
Substitute –6, 90, 15 for x, y and z respectively in
then solve for k.
Substitute 9, –5 and 1 for x, z and k respectively then
evaluate.
ANSWER: ANSWER: –45
85. SHOPPING A certain store found that the number
of customers who will attend a sale can be modeled
by
where N is the number of
customers expected, P is the percent of the sale
discount, and t is the number of hours the sale will
last. Find the number of customers the store should
expect for a sale that is 50% off and will last four
hours.
84. Suppose y varies jointly as x and z. Find y when x =
9 and z = –5, if y = –90 when z = 15 and x = –6.
SOLUTION: Given P = 50% or 0.5 and t = 4.
SOLUTION: Given
.
Find the value of k.
Substitute –6, 90, 15 for x, y and z respectively in
eSolutions Manual
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then solve for k.
ANSWER: 731 customers
Page 31
Evaluate each expression if a = –2, b =
and ANSWER: ANSWER: 10-3–45
Geometric Sequences and Series
85. SHOPPING A certain store found that the number
of customers who will attend a sale can be modeled
by
where N is the number of
customers expected, P is the percent of the sale
discount, and t is the number of hours the sale will
last. Find the number of customers the store should
expect for a sale that is 50% off and will last four
hours.
87. SOLUTION: Substitute the values of the variables.
SOLUTION: Given P = 50% or 0.5 and t = 4.
ANSWER: ANSWER: 731 customers
88. Evaluate each expression if a = –2, b =
and c = –12.
SOLUTION: Substitute the values of the variables.
86. SOLUTION: Substitute the values of the variables.
ANSWER: 36
ANSWER: 89. SOLUTION: Substitute the values of the variables.
87. eSolutions Manual - Powered by Cognero
SOLUTION: Substitute the values of the variables.
Page 32
ANSWER: 10-336
Geometric Sequences and Series
89. SOLUTION: Substitute the values of the variables.
ANSWER: eSolutions Manual - Powered by Cognero
Page 33
10-4 Infinite Geometric Series
ANSWER: divergent
Determine whether each infinite geometric
series is convergent or divergent.
3. 0.5 + 0.7 + 0.98 + …
SOLUTION: Find the value of r.
1. 16 – 8 + 4 – …
SOLUTION: Find the value of r.
Since
Since
, the series is divergent.
, the series is convergent.
ANSWER: divergent
ANSWER: convergent
4. 1 + 1 + 1 + …
2. 32 – 48 + 72 – …
SOLUTION: Find the value of r.
SOLUTION: Find the value of r.
Since 1 = 1 , the series is divergent.
Since
, the series is divergent.
ANSWER: divergent
ANSWER: divergent
Find the sum of each infinite series, if it exists.
5. 440 + 220 + 110 + …
3. 0.5 + 0.7 + 0.98 + …
SOLUTION: Find the value of r.
SOLUTION: Find the value of r.
Since
Since
, the series is divergent.
ANSWER: divergent
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4. 1 + 1 + 1 + …
, the series is convergent.
Find the sum.
Page 1
ANSWER: 880
ANSWER: 10-4divergent
Infinite Geometric Series
Find the sum of each infinite series, if it exists.
6. 520 + 130 + 32.5 + …
5. 440 + 220 + 110 + …
SOLUTION: Find the value of r.
SOLUTION: Find the value of r.
Since
Since
, the series is convergent.
, the series is convergent.
Find the sum.
Find the sum.
440 + 220 + 110 + … = 880
520 + 130 + 32.5 + … =
ANSWER: 880
ANSWER: 6. 520 + 130 + 32.5 + …
7. SOLUTION: Find the value of r.
SOLUTION: Find the value of r.
Since
, the series is convergent.
Find the sum.
Since
, the series is diverges and the sum
does not exist.
eSolutions Manual - Powered by Cognero
ANSWER: No sum exists.
Page 2
ANSWER: ANSWER: No sum exists.
10-4 Infinite Geometric Series
9. CCSS SENSE-MAKING A certain drug has a
half-life of 8 hours after it is administered to a
patient. What percent of the drug is still in the
patient’s system after 24 hours?
7. SOLUTION: Find the value of r.
SOLUTION: Given a 1 = 100% or 1 and r =
and n = 4.
Since
, the series is diverges and the sum
does not exist.
ANSWER: No sum exists.
After 24 hours, 12.5% of the drug is still in the
patient’s system.
8. ANSWER: 12.5%
SOLUTION: Find the value of r.
Find the sum of each infinite series, if it exists.
10. SOLUTION: Since
Since
, the series is diverges and the
sum does not exists.
, the series is diverges and the sum
does not exist.
ANSWER: No sum exists.
ANSWER: No sum exists.
9. CCSS SENSE-MAKING A certain drug has a
half-life of 8 hours after it is administered to a
patient. What percent of the drug is still in the
patient’s system after 24 hours?
Given a = 100% or 1 and r =
SOLUTION: Since
SOLUTION: 1 - Powered by Cognero
eSolutions Manual
11. , the series is convergent.
and n = 4.
Page 3
ANSWER: No sum exists.
10-4 Infinite Geometric Series
11. ANSWER: 15
13. SOLUTION: SOLUTION: Since
, the series is convergent.
Since
, the series is convergent.
ANSWER: –4
ANSWER: 2
12. Write each repeating decimal as a fraction.
SOLUTION: Since
, the series is convergent.
14. SOLUTION: Find the value of r.
ANSWER: 15
13. eSolutions Manual - Powered by Cognero
SOLUTION: Page 4
ANSWER: 10-42Infinite Geometric Series
Write each repeating decimal as a fraction.
14. 15. SOLUTION: SOLUTION: Find the value of r.
Find the value of r.
ANSWER: ANSWER: Determine whether each infinite geometric
series is convergent or divergent.
16. 21 + 63 + 189 + …
SOLUTION: Find the value of r.
15. SOLUTION: Since
, the series is divergent.
Find the value of r.
eSolutions Manual - Powered by Cognero
ANSWER: divergent
Page 5
ANSWER: ANSWER: convergent
10-4 Infinite Geometric Series
Determine whether each infinite geometric
series is convergent or divergent.
18. 16. 21 + 63 + 189 + …
SOLUTION: Find the value of r.
SOLUTION: Find the value of r.
Since
, the series is divergent.
Since
ANSWER: divergent
, the series is divergent.
ANSWER: divergent
17. 480 + 360 + 270 + …
19. SOLUTION: Find the value of r.
SOLUTION: Find the value of r.
Since
, the series is convergent.
ANSWER: convergent
Since
, the series is divergent.
ANSWER: divergent
18. SOLUTION: Find the value of r.
20. 0.1 + 0.01 + 0.001 + …
SOLUTION: Find the value of r.
Since
, the series is divergent.
eSolutions Manual - Powered by Cognero
Since
ANSWER: , the series is convergent.
Page 6
ANSWER: 10-4divergent
Infinite Geometric Series
ANSWER: divergent
20. 0.1 + 0.01 + 0.001 + …
Find the sum of each infinite series, if it exists.
SOLUTION: Find the value of r.
22. 18 + 21.6 + 25.92 + …
SOLUTION: Find the value of r.
Since
, the series is convergent.
Since
, the series is diverges and the sum
does not exist.
ANSWER: convergent
ANSWER: No sum exists.
21. 0.008 + 0.08 + 0.8 + …
SOLUTION: Find the value of r.
23. –3 – 4.2 – 5.88 – …
SOLUTION: Find the value of r.
Since
, the series is divergent.
ANSWER: divergent
Since
, the series is diverges and the sum
does not exists.
Find the sum of each infinite series, if it exists.
ANSWER: No sum exists.
22. 18 + 21.6 + 25.92 + …
SOLUTION: Find the value of r.
Since
, the series is diverges and the sum
does not exist.
24. SOLUTION: Find the value of r.
ANSWER: No sum exists.
Since
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23. –3 – 4.2 – 5.88 – …
, the series is convergent.
Page 7
Find the sum.
ANSWER: ANSWER: sum exists.
10-4No
Infinite
Geometric Series
24. 25. SOLUTION: Find the value of r.
SOLUTION: Find the value of r.
Since
, the series is convergent.
Since
Find the sum.
Find the sum.
ANSWER: ANSWER: 25. , the series is convergent.
26. SOLUTION: Find the value of r.
SOLUTION: Find the value of r.
Since
Since
, the series is convergent.
eSolutions Manual - Powered by Cognero
Find the sum.
, the series is convergent.
Find the sum.
Page 8
ANSWER: ANSWER: No sum exists.
10-4 Infinite Geometric Series
26. SOLUTION: Find the value of r.
28. SWINGS If Kerry does not push any harder after
his initial swing, the distance traveled per swing will
decrease by 10% with each swing. If his initial swing
traveled 6 feet, find the total distance traveled when
he comes to rest.
Since
, the series is convergent.
Find the sum.
SOLUTION: Given a 1 = 6 and r = 100% – 10% or 0.9
Find the sum.
The total distance traveled is 60 ft.
ANSWER: 63
ANSWER: 60 ft
27. 32 + 40 + 50 + …
SOLUTION: Find the value of r.
Find the sum of each infinite series, if it exists.
29. SOLUTION: Since
, the series is diverges and the sum
does not exists.
Since
sum does not exists.
ANSWER: No sum exists.
28. SWINGS If Kerry does not push any harder after
his initial swing, the distance traveled per swing will
decrease
by- Powered
10% with
each swing. If his initial swing
eSolutions
Manual
by Cognero
traveled 6 feet, find the total distance traveled when
he comes to rest.
, the series is diverges and the
ANSWER: No sum exists.
Page 9
30. ANSWER: ft
10-460
Infinite
Geometric Series
ANSWER: No sum exists.
Find the sum of each infinite series, if it exists.
31. 29. SOLUTION: SOLUTION: Since
Since
, the series is convergent.
, the series is diverges and the
sum does not exists.
ANSWER: No sum exists.
30. SOLUTION: Since
does not exists.
, the series is diverges and the sum
ANSWER: ANSWER: No sum exists.
32. 31. SOLUTION: Since
SOLUTION: Since
, the series is diverges and the
sum does not exists.
, the series is convergent.
ANSWER: No sum exists.
33. SOLUTION: Since
eSolutions Manual - Powered by Cognero
, the series is convergent.
Page 10
ANSWER: sum exists.
10-4No
Infinite
Geometric Series
ANSWER: 16
33. 34. SOLUTION: SOLUTION: Since
, the series is convergent.
Since
, the series is convergent.
ANSWER: 16
ANSWER: 34. Write each repeating decimal as a fraction.
SOLUTION: Since
35. , the series is convergent.
SOLUTION: The number
can be written as .
Find the value of r.
ANSWER: eSolutions Manual - Powered by Cognero
Page 11
ANSWER: ANSWER: 10-4 Infinite Geometric Series
Write each repeating decimal as a fraction.
36. 35. SOLUTION: The number
can be written as The number
SOLUTION: .
can be written as .
Find the value of r.
Find the value of r.
Therefore:
Therefore:
ANSWER: ANSWER: 37. 36. SOLUTION: The number
can be written as SOLUTION: The number
.
can be written as .
eSolutions Manual - Powered by Cognero
Find the value of r.
Page 12
ANSWER: ANSWER: 10-4 Infinite Geometric Series
38. 37. SOLUTION: SOLUTION: The number
can be written as The number
can be written as .
.
Find the value of r.
Find the value of r.
Therefore:
Therefore:
ANSWER: ANSWER: 38. 39. SOLUTION: SOLUTION: The number
can be written as The number
.
can be written as .
eSolutions Manual - Powered by Cognero
Page 13
Find the value of r.
Find the value of r.
ANSWER: ANSWER: 10-4 Infinite Geometric Series
39. 40. SOLUTION: SOLUTION: The number
can be written as The number
can be written as
.
.
Find the value of r.
Find the value of r.
Therefore:
Therefore:
ANSWER: ANSWER: 41. FANS A fan is running at 10 revolutions per second.
After it is turned off, its speed decreases at a rate of
75% per second. Determine the number of
revolutions completed by the fan after it is turned off.
40. SOLUTION: The number
can be written as
.
SOLUTION: Given a 1 = 10 and r = 100% – 75% or 0.25.
eSolutions Manual - Powered by Cognero
Find the value of r.
Find the sum.
Page 14
ANSWER: ANSWER: 10-4 Infinite Geometric Series
41. FANS A fan is running at 10 revolutions per second.
After it is turned off, its speed decreases at a rate of
75% per second. Determine the number of
revolutions completed by the fan after it is turned off.
42. CCSS PRECISION Kamiko deposited $5000 into
an account at the beginning of the year. The account
earns 8% interest each year.
a. How much money will be in the account after 20
SOLUTION: Given a 1 = 10 and r = 100% – 75% or 0.25.
years? (Hint: Let 5000(1 + 0.08) represent the end
of the first year.)
b. Is this series convergent or divergent? Explain.
Find the sum.
1
SOLUTION: Given principal = 5000, interest 8% or 0.08 years =
20.
a.
b. It is a diverging series. The ratio is 1.08, which is
greater than 1.
The fan completed
revolutions after it is turned ANSWER: a. $23,304.79
off.
b. It is a diverging series. The ratio is 1.08, which is
greater than 1.
ANSWER: 42. CCSS PRECISION Kamiko deposited $5000 into
an account at the beginning of the year. The account
earns 8% interest each year.
a. How much money will be in the account after 20
1
43. RECHARGEABLE BATTERIES A certain
rechargeable battery is advertised to recharge back
to 99.9% of its previous capacity with every charge.
If its initial capacity is 8 hours of life, how many total
hours should the battery last?
SOLUTION: Given a 1 = 8 and r = 99.9% or 0.999
years? (Hint: Let 5000(1 + 0.08) represent the end
of the first year.)
Find the sum.
b. Is this series convergent or divergent? Explain.
SOLUTION: Given principal = 5000, interest 8% or 0.08 years =
20.
a.
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b. It is a diverging series. The ratio is 1.08, which is
ANSWER: 8000 hrs
Page 15
Find the sum of each infinite series, if it exists.
a. $23,304.79
ANSWER: 8000 hrs
b. It is a diverging series. The ratio is 1.08, which is
than
1.
10-4greater
Infinite
Geometric
Series
43. RECHARGEABLE BATTERIES A certain
rechargeable battery is advertised to recharge back
to 99.9% of its previous capacity with every charge.
If its initial capacity is 8 hours of life, how many total
hours should the battery last?
Find the sum of each infinite series, if it exists.
44. SOLUTION: Find the value of r.
SOLUTION: Given a 1 = 8 and r = 99.9% or 0.999
Find the sum.
Since
, the series is convergent.
Find the sum.
ANSWER: 8000 hrs
Find the sum of each infinite series, if it exists.
44. SOLUTION: Find the value of r.
ANSWER: Since
, the series is convergent.
Find the sum.
45. SOLUTION: Find the value of r.
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Since
Page 16
, the series is convergent.
ANSWER: ANSWER: 10-4 Infinite Geometric Series
45. 46. SOLUTION: Find the value of r.
SOLUTION: Find the value of r.
Since
, the series is convergent.
Since
Find the sum.
Find the sum.
ANSWER: ANSWER: 46. , the series is convergent.
47. SOLUTION: Find the value of r.
SOLUTION: Find the value of r.
Since
, the series is convergent.
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Since
, the series is diverges and the sum
does not exists.
Find the sum.
Page 17
ANSWER: ANSWER: No sum exists.
10-4 Infinite Geometric Series
49. 47. SOLUTION: Find the value of r.
SOLUTION: Find the value of r.
Since
, the series is diverges and the sum
Since
does not exists.
Find the sum.
ANSWER: No sum exists.
, the series is convergent.
48. SOLUTION: Find the value of r.
ANSWER: Since
, the series is diverges and the sum
does not exists.
ANSWER: No sum exists.
49. SOLUTION: Find the value of r.
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50. MULTIPLE REPRESENTATIONS In this
problem, you will use a square of paper that is at
least 8 inches on a side.
a. CONCRETE Let the square be one unit. Cut
away one half of the square. Call this piece Term 1.
Next, cut away one half of the remaining sheet of
paper. Call this piece Term 2. Continue cutting the
remaining paper in half and labeling the pieces with a
term number as long as possible. List the fractions
represented by the pieces.
b. NUMERICAL If you could cut the squares
indefinitely, you would have an infinite series. Find
the sum of the series.
Page 18
c. VERBAL How does the sum of the series relate
remaining paper in half and labeling the pieces with a
term number as long as possible. List the fractions
represented by the pieces.
10-4 Infinite Geometric Series
b. NUMERICAL If you could cut the squares
indefinitely, you would have an infinite series. Find
the sum of the series.
c. VERBAL How does the sum of the series relate
to the original square of paper?
b. 1
c. The original square has area 1 unit and the area of
all the pieces cannot exceed 1.
51. PHYSICS In a physics experiment, a steel ball on a
flat track is accelerated, and then allowed to roll
freely. After the first minute, the ball has rolled 120
feet. Each minute the ball travels only 40% as far as
it did during the preceding minute. How far does the
ball travel?
SOLUTION: a.
SOLUTION: Given a 1 = 120 and r = 40% or 0.4.
b. Find r.
Find the sum.
Find the sum.
The ball travels 200 ft.
ANSWER: 200 ft
c. The original square has area 1 unit and the area of
all the pieces cannot exceed 1.
ANSWER: a.
52. PENDULUMS A pendulum travels 12 centimeters
on its first swing and 95% of that distance on each
swing thereafter. Find the total distance traveled by
the pendulum when it comes to rest.
SOLUTION: Given a 1 = 12 and r = 95% or 0.95
b. 1
Find the sum.
c. The original square has area 1 unit and the area of
all the pieces cannot exceed 1.
51. PHYSICS In a physics experiment, a steel ball on a
flat track is accelerated, and then allowed to roll
freely. After the first minute, the ball has rolled 120
feet. Each minute the ball travels only 40% as far as
it did during the preceding minute. How far does the
ball travel?
SOLUTION: Given a 1 = 120 and r = 40% or 0.4.
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Find the sum.
The pendulum travels 240 cm.
ANSWER: 240 cm
19
of its
53. TOYS If a rubber ball can bounce back to 95%Page
original height, what is the total vertical distance that
it will travel if it is dropped from an elevation of 30
ANSWER: 10-4240
Infinite
cm Geometric Series
53. TOYS If a rubber ball can bounce back to 95% of its
original height, what is the total vertical distance that
it will travel if it is dropped from an elevation of 30
feet?
SOLUTION: Distance traveled by the rubber ball in downward
direction.
ANSWER: 1170 ft
54. CARS During a maintenance inspection, a tire is
removed from a car and spun on a diagnostic
machine. When the machine is turned off, the
spinning tire completes 20 revolutions the first second
and 98% of the revolutions each additional second.
How many revolutions does the tire complete before
it stops spinning?
Given, a 1 = 30 and r = 95% or 0.95.
SOLUTION: Given a 1 = 20 and r = 98% or 0.98
Find the sum.
Find the sum.
Distance traveled by the rubber ball in upward
direction.
The tire completes 1000 revolutions.
Given, a 1 = 28.5 and r = 95% or 0.95.
Find the sum.
The total distance traveled by the rubber ball is 600 +
570 or 1170 ft.
ANSWER: 1170 ft
54. CARS During a maintenance inspection, a tire is
removed from a car and spun on a diagnostic
machine. When the machine is turned off, the
spinning tire completes 20 revolutions the first second
and 98% of the revolutions each additional second.
How many revolutions does the tire complete before
it stops spinning?
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SOLUTION: Given a 1 = 20 and r = 98% or 0.98
ANSWER: 1000 revolutions
55. ECONOMICS A state government decides to
stimulate its economy by giving $500 to every adult.
The government assumes that everyone who
receives the money will spend 80% on consumer
goods and that the producers of these goods will in
turn spend 80% on consumer goods. How much
money is generated for the economy for every $500
that the government provides?
SOLUTION: Here, a 1 = 500 and r = 80% or 0.8.
Find the sum.
ANSWER: $2500
Page 20
ANSWER: revolutions
10-41000
Infinite
Geometric Series
ANSWER: $2500
55. ECONOMICS A state government decides to
stimulate its economy by giving $500 to every adult.
The government assumes that everyone who
receives the money will spend 80% on consumer
goods and that the producers of these goods will in
turn spend 80% on consumer goods. How much
money is generated for the economy for every $500
that the government provides?
56. SCIENCE MUSEUM An exhibit at a science
museum offers visitors the opportunity to experiment
with the motion of an object on a spring. One visitor
pulls the object down and lets it go. The object
travels 1.2 feet upward before heading back the
other way. Each time the object changes direction, it
decreases its distance by 20% when compared to the
previous direction. Find the total distance traveled by
the object.
SOLUTION: Here, a 1 = 500 and r = 80% or 0.8.
SOLUTION: Here a 1 = 1.2, r = 1 – 0.2 or 0.8.
Find the sum.
ANSWER: $2500
The total distance traveled by the object is 6 feet.
ANSWER: 6 ft
56. SCIENCE MUSEUM An exhibit at a science
museum offers visitors the opportunity to experiment
with the motion of an object on a spring. One visitor
pulls the object down and lets it go. The object
travels 1.2 feet upward before heading back the
other way. Each time the object changes direction, it
decreases its distance by 20% when compared to the
previous direction. Find the total distance traveled by
the object.
Match each graph with its corresponding
description.
SOLUTION: Here a 1 = 1.2, r = 1 – 0.2 or 0.8.
57. a. converging geometric series
b. diverging geometric series
c. converging arithmetic series
d. diverging arithmetic series
The total distance traveled by the object is 6 feet.
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ANSWER: 6 ft
SOLUTION: Page 21
The graph is diverging geometric series. Therefore,
option b is the correct answer.
ANSWER: b
ANSWER: ft
10-46Infinite
Geometric Series
Match each graph with its corresponding
description.
58. a. converging geometric series
57. b. diverging geometric series
a. converging geometric series
c. converging arithmetic series
b. diverging geometric series
d. diverging arithmetic series
c. converging arithmetic series
SOLUTION: The graph is diverging arithmetic series. Therefore,
option d is the correct answer.
d. diverging arithmetic series
SOLUTION: The graph is diverging geometric series. Therefore,
option b is the correct answer.
ANSWER: d
ANSWER: b
59. a. converging geometric series
58. a. converging geometric series
b. diverging geometric series
b. diverging geometric series
c. converging arithmetic series
d. diverging arithmetic series
c. converging arithmetic series
SOLUTION: The graph is converging geometric series. Therefore,
option a is the correct answer.
d. diverging arithmetic series
SOLUTION: The graph is diverging arithmetic series. Therefore,
option
d is-the
correct
answer.
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ANSWER: ANSWER: a
Page 22
60. ERROR ANALYSIS Emmitt and Austin are asked
ANSWER: 10-4dInfinite Geometric Series
option a is the correct answer.
ANSWER: a
60. ERROR ANALYSIS Emmitt and Austin are asked
to find the sum of 1 – 1 + 1 – … . Is either of them
correct? Explain your reasoning.
59. a. converging geometric series
b. diverging geometric series
c. converging arithmetic series
d. diverging arithmetic series
SOLUTION: The graph is converging geometric series. Therefore,
option a is the correct answer.
ANSWER: a
60. ERROR ANALYSIS Emmitt and Austin are asked
to find the sum of 1 – 1 + 1 – … . Is either of them
correct? Explain your reasoning.
SOLUTION: Sample answer: Austin; the common ratio of the
series r = –1, so the absolute value of r = 1 and the
series diverges.
ANSWER: Sample answer: Austin; the common ratio of the
series r = –1, so the absolute value of r = 1 and the
series diverges.
61. PROOF Derive the formula for the sum of an
infinite geometric series.
SOLUTION: Sample answer:
The sum of a geometric series is
.
SOLUTION: Sample answer: Austin; the common ratio of the
series r = –1, so the absolute value of r = 1 and the
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series
diverges.
For an infinite series with | r | < 1,
.
Thus,
ANSWER: Sample answer:
Page 23
ANSWER: Sample answer: Austin; the common ratio of the
series r = –1, so the absolute value of r = 1 and the
diverges.
10-4series
Infinite
Geometric Series
Thus,
61. PROOF Derive the formula for the sum of an
infinite geometric series.
62. CHALLENGE For what values of b does 3 + 9b +
2
3
27b + 81b + … have a sum?
SOLUTION: Sample answer:
SOLUTION: The common ratio is
The sum of a geometric series is
.
.
The series have a sum, if the absolute value of the
common ratio is less than 1.
For an infinite series with | r | < 1,
.
That is,
.
Thus,
Solve the inequality.
ANSWER: Sample answer:
The sum of a geometric series is
.
Therefore, the value of b should be
For an infinite series with | r | < 1,
.
Thus,
ANSWER: .
62. CHALLENGE For what values of b does 3 + 9b +
2
3
27b + 81b + … have a sum?
SOLUTION: 63. REASONING When does an infinite geometric
series have a sum, and when does it not have a sum?
Explain your reasoning.
The common ratio is
.
The series have a sum, if the absolute value of the
common ratio is less than 1.
That is,
.
Solve the inequality.
SOLUTION: Sample answer: An infinite geometric series has a
sum when the common ratio has an absolute value
less than 1. When this occurs, the terms will
approach 0 as n approaches infinity. With the future
terms almost 0, the sum of the series will approach a
limit. When the common ratio is 1 or greater, the
terms will keep increasing and approach infinity as n
approaches infinity and the sum of the series will
have no limit.
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Therefore, the value of b should be
.
ANSWER: Sample answer: An infinite geometric series has a
sum when the common ratio has an absolute value
less than 1. When this occurs, the terms will
Page 24
approach 0 as n approaches infinity. With the future
terms almost 0, the sum of the series will approach a
limit. When the common ratio is 1 or greater, the
ANSWER: 10-4 Infinite Geometric Series
63. REASONING When does an infinite geometric
series have a sum, and when does it not have a sum?
Explain your reasoning.
SOLUTION: Sample answer: An infinite geometric series has a
sum when the common ratio has an absolute value
less than 1. When this occurs, the terms will
approach 0 as n approaches infinity. With the future
terms almost 0, the sum of the series will approach a
limit. When the common ratio is 1 or greater, the
terms will keep increasing and approach infinity as n
approaches infinity and the sum of the series will
have no limit.
ANSWER: Sample answer: An infinite geometric series has a
sum when the common ratio has an absolute value
less than 1. When this occurs, the terms will
approach 0 as n approaches infinity. With the future
terms almost 0, the sum of the series will approach a
limit. When the common ratio is 1 or greater, the
terms will keep increasing and approach infinity as n
approaches infinity and the sum of the series will
have no limit.
64. CCSS ARGUMENTS Determine whether the
following statement is sometimes, always, or never
true. Explain your reasoning.
If the absolute value of a term of any geometric
series is greater than the absolute value of the
previous term, then the series is divergent.
SOLUTION: Sample answer: Sometimes; the statement is true for
all infinite geometric series.
ANSWER: Sample answer: Sometimes; the statement is true for
all infinite geometric series.
65. OPEN ENDED Write an infinite series with a sum
that converges to 9.
SOLUTION: Sample answer:
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ANSWER: Sample answer: Sometimes; the statement is true for
all infinite geometric series.
65. OPEN ENDED Write an infinite series with a sum
that converges to 9.
SOLUTION: Sample answer:
ANSWER: Sample answer:
66. OPEN ENDED Write 3 – 6 + 12 – … using sigma
notation in two different ways.
SOLUTION: ANSWER: 67. WRITING IN MATH Explain why an arithmetic
series is always divergent.
SOLUTION: An arithmetic series has a common difference, so
each term will eventually become more positive or
more negative, but never approach 0. With the terms
not approaching 0, the sum will never reach a limit
and the series cannot converge.
ANSWER: An arithmetic series has a common difference, so
each term will eventually become more positive or
more negative, but never approach 0. With the terms
not approaching 0, the sum will never reach a limit
and the series cannot converge.
Page 25
68. SAT/ACT What is the sum of an infinite geometric
series with a first term of 27 and a common ratio of
ANSWER: 10-4 Infinite Geometric Series
67. WRITING IN MATH Explain why an arithmetic
series is always divergent.
each term will eventually become more positive or
more negative, but never approach 0. With the terms
not approaching 0, the sum will never reach a limit
and the series cannot converge.
68. SAT/ACT What is the sum of an infinite geometric
series with a first term of 27 and a common ratio of
SOLUTION: An arithmetic series has a common difference, so
each term will eventually become more positive or
more negative, but never approach 0. With the terms
not approaching 0, the sum will never reach a limit
and the series cannot converge.
ANSWER: An arithmetic series has a common difference, so
each term will eventually become more positive or
more negative, but never approach 0. With the terms
not approaching 0, the sum will never reach a limit
and the series cannot converge.
A 18
B 34
C 41
D 65
E 81
SOLUTION: 68. SAT/ACT What is the sum of an infinite geometric
series with a first term of 27 and a common ratio of
Given a 1 = 27, r =
.
Find the sum.
A 18
B 34
C 41
D 65
Option E is the correct answer.
ANSWER: E
69. Adelina, Michelle, Masao, and Brandon each
simplified the same expression at the board. Each
student’s work is shown below. The teacher said that
while two of them had a correct answer, only one of
them had arrived at the correct conclusion using
correct steps.
Adelina’s work
Masao’s work
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Page 26
Adelina’s work
10-4 Infinite Geometric Series
Masao’s work
ANSWER: H
70. GRIDDED RESPONSE Evaluate log8 60 to the
nearest hundredth.
SOLUTION: Michelle’s work
Brandon’s work
ANSWER: 1.97
71. GEOMETRY The radius of a large sphere was
Which is a completely accurate simplification?
multiplied by a factor of
to produce a smaller sphere.
F Adelina’s work
G Michelle’s work
H Masao’s work
J Brandon’s work
How does the volume of the smaller sphere compare
to the volume of the larger sphere?
SOLUTION: Masao’s work is a completely accurate
simplification.
Option H is the correct answer.
A The volume of the smaller sphere is
B The volume of the smaller sphere is
ANSWER: H
70. GRIDDED RESPONSE Evaluate log8 60 to the
as large.
C The volume of the smaller sphere is
as large.
as large.
nearest hundredth.
D The volume of the smaller sphere is
SOLUTION: SOLUTION: The volume of the smaller sphere is
as large.
as large.
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Option C is the correct answer.
ANSWER: C
71. GEOMETRY The radius of a large sphere was
Page 27
ANSWER: 10-41.97
Infinite Geometric Series
ANSWER: C
71. GEOMETRY The radius of a large sphere was
multiplied by a factor of
to produce a smaller sphere.
72. GAMES An audition is held for a TV game show.
At the end of each round, one half of the prospective
contestants are eliminated from the competition. On
a particular day, 524 contestants begin the audition.
a. Write an equation for finding the number of
contestants who are left after n rounds.
b. Using this method, will the number of contestants
who are to be eliminated always be a whole number?
Explain.
How does the volume of the smaller sphere compare
to the volume of the larger sphere?
A The volume of the smaller sphere is
SOLUTION: a. Given a 10 = 524 and r =
as large.
.
B The volume of the smaller sphere is
as large.
C The volume of the smaller sphere is
as large.
D The volume of the smaller sphere is
b. No; at the beginning of the third round there will
be 131 contestants and one-half of that is 65.5.
as large.
ANSWER: a.
SOLUTION: The volume of the smaller sphere is
as large.
b. No; at the beginning of the third round there will
be 131 contestants and one-half of that is 65.5.
Option C is the correct answer.
ANSWER: C
72. GAMES An audition is held for a TV game show.
At the end of each round, one half of the prospective
contestants are eliminated from the competition. On
a particular day, 524 contestants begin the audition.
a. Write an equation for finding the number of
contestants who are left after n rounds.
b. Using this method, will the number of contestants
who are to be eliminated always be a whole number?
Explain.
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SOLUTION: 73. CLUBS A quilting club consists of 9 members.
Every week, each member must bring one completed
quilt square.
a. Find the first eight terms of the sequence that
describes the total number of squares that have been
made after each meeting.
b. One particular quilt measures 72 inches by 84
inches and is being designed with 4-inch squares.
After how many meetings will the quilt be complete?
SOLUTION: a. 9, 18, 27, 36, 45, 54, 63, 72
b. The number of squares required is
.
Page 28
ANSWER: a. 9, 18, 27, 36, 45, 54, 63, 72
b. No; at the beginning of the third round there will
131 contestants
andSeries
one-half of that is 65.5.
10-4be
Infinite
Geometric
b. 42 meetings
73. CLUBS A quilting club consists of 9 members.
Every week, each member must bring one completed
quilt square.
Find each function value.
74. f (x) = 5x – 9, f (6) a. Find the first eight terms of the sequence that
describes the total number of squares that have been
made after each meeting.
SOLUTION: Substitute 6 for x and evaluate.
b. One particular quilt measures 72 inches by 84
inches and is being designed with 4-inch squares.
After how many meetings will the quilt be complete?
SOLUTION: a. 9, 18, 27, 36, 45, 54, 63, 72
ANSWER: 21
b. The number of squares required is
.
That is a n = 378.
Given a 1 = 9 and d = 9.
2
75. g(x) = x – x, g(4)
SOLUTION: Substitute 4 for x and evaluate.
Find n.
ANSWER: 12
2
76. h(x) = x – 2x – 1, h(3)
ANSWER: a. 9, 18, 27, 36, 45, 54, 63, 72
b. 42 meetings
SOLUTION: Substitute 3 for x and evaluate.
Find each function value.
74. f (x) = 5x – 9, f (6) SOLUTION: Substitute 6 for x and evaluate.
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ANSWER: 2
Page 29
10-5 Recursion and Iteration
ANSWER: –3, 5, 13, 21, 29
Find the first five terms of each sequence
described.
1. a 1 = 16, a n + 1 = a n + 4
3. a 1 = 5, a n + 1 = 3a n + 2
SOLUTION: SOLUTION: The first five terms of the sequence are 16, 20, 24,
28 and 32.
ANSWER: 16, 20, 24, 28, 32
2. a 1 = –3, a n + 1 = a n + 8
The first five terms of the sequence are 5, 17, 53,
161 and 485.
ANSWER: 5, 17, 53, 161, 485
4. a 1 = –4, a n + 1 = 2a n – 6
SOLUTION: SOLUTION: The first five terms of the sequence are –3, 5, 13, 21
and 29.
ANSWER: –3, 5, 13, 21, 29
3. a 1 = 5, a n + 1 = 3a n + 2
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The first five terms of the sequence are –4, –14, –34,
–74 and –154.
ANSWER: –4, –14, –34, –74, –154
Write a recursive formula for each sequence.
5. 3, 8, 18, 38, 78, …
SOLUTION: Page 1
ANSWER: –14, –34,and
–74,Iteration
–154
10-5–4,
Recursion
a. Write a recursive formula for the balance owed at
the end of each month.
b. Find the balance owed after the first four months.
Write a recursive formula for each sequence.
c. How much interest has accumulated after the first
six months?
5. 3, 8, 18, 38, 78, …
SOLUTION: SOLUTION: a. Here a 1 = 1500.
The recursive formula for the balance owed at the
end of each month is
.
b. Substitute n = 1 in
.
ANSWER: a n + 1 = 2a n + 2; a 1 = 3
6. 5, 14, 41, 122, 365, …
SOLUTION: The balance owed after the first four months will be
$1154.87.
c. Find the balance owed after the first six months.
ANSWER: a n + 1 = 3a n – 1; a 1 = 5
7. FINANCING Ben financed a $1500 rowing
machine to help him train for the college rowing
team. He could only make a $100 payment each
month, and his bill increased by 1% due to interest at
the end of each month.
a. Write a recursive formula for the balance owed at
the end of each month.
b. Find the balance owed after the first four months.
c. How much interest has accumulated after the first
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six months?
Amount paid for the first six months is $600.
So, after six months $900 should be paid if there is no
interest at all.
After the first six months, the interest has
accumulated to $77.08.
ANSWER: a. a n = 1.01a n – 1 – 100; a 1 = 1500
b. $1415, $1329.15, $1242.44, $1154.87
c. $77.08
Page 2
Find the first three iterates of each function for
ANSWER: a. a n = 1.01a n – 1 – 100; a 1 = 1500
10-5 Recursion and Iteration
b. $1415, $1329.15, $1242.44, $1154.87
c. $77.08
Find the first three iterates of each function for
the given initial value.
ANSWER: –18, 74, –294
10. f (x) = 6x + 3, x 0 = –4
SOLUTION: 8. f (x) = 5x + 2, x 0 = 8
SOLUTION: The first three iterates are –21, –123 and –735.
ANSWER: –21, –123, –735
The first three iterates are 42, 212 and 1062.
ANSWER: 42, 212, 1062
11. f (x) = 8x – 4, x0 = –6
SOLUTION: 9. f (x) = –4x + 2, x0 = 5
SOLUTION: The first three iterates are –52, –420 and –3364.
ANSWER: –52, –420, –3364
The first three iterates are –18, 74 and –294.
ANSWER: –18, 74, –294
10. f (x) = 6x + 3, x 0 = –4
CCSS PERSEVERANCE Find the first five
terms of each sequence described.
12. a 1 = 10, a n + 1 = 4a n + 1
SOLUTION: SOLUTION: eSolutions Manual - Powered by Cognero
Page 3
ANSWER: –420, –3364
10-5–52,
Recursion
and Iteration
CCSS PERSEVERANCE Find the first five
terms of each sequence described.
12. a 1 = 10, a n + 1 = 4a n + 1
ANSWER: –9, –10, –12, –16, –24
14. a 1 = 12, a n + 1 = a n + n
SOLUTION: SOLUTION: The first five terms of the sequence are 12, 13, 15,
18 and 22.
The first five terms of the sequence are 10, 41, 165,
661 and 2645.
ANSWER: 10, 41, 165, 661, 2645
13. a 1 = –9, a n + 1 = 2a n + 8
ANSWER: 12, 13, 15, 18, 22
15. a 1 = –4, a n + 1 = 2a n + n
SOLUTION: SOLUTION: The first five terms of the sequence are –4, –7, –12,
–21 and –38.
The first five terms of the sequence are –9, –10, –12,
–16 and –24.
ANSWER: –9, –10, –12, –16, –24
14. a 1 = 12, a n + 1 = a n + n
ANSWER: –4, –7, –12, –21, –38
16. a 1 = 6, a n + 1 = 3a n – n
SOLUTION: SOLUTION: eSolutions Manual - Powered by Cognero
Page 4
ANSWER: –7, –12, –21,
10-5–4,
Recursion
and –38
Iteration
16. a 1 = 6, a n + 1 = 3a n – n
ANSWER: –2, –8, –36, –174, –862
18. a 1 = 7, a 2 = 10, a n + 2 = 2a n + a n + 1
SOLUTION: SOLUTION: The first five terms of the sequence are 6, 17, 49,
144 and 428.
ANSWER: 6, 17, 49, 144, 428
17. a 1 = –2, a n + 1 = 5a n + 2n
The first five terms of the sequence are 7, 10, 24, 44
and 92.
ANSWER: 7, 10, 24, 44, 92
19. a 1 = 4, a 2 = 5, a n + 2 = 4a n – 2a n + 1
SOLUTION: SOLUTION: The first five terms of the sequence are 4, 5, 6, 8 and
8.
The first five terms of the sequence are –2, –8, –36,
–174 and –862.
ANSWER: 4, 5, 6, 8, 8
ANSWER: –2, –8, –36, –174, –862
18. a 1 = 7, a 2 = 10, a n + 2 = 2a n + a n + 1
20. a 1 = 4, a 2 = 3x, an = a n – 1 + 4a n – 2
SOLUTION: SOLUTION: eSolutions Manual - Powered by Cognero
The first five terms of the sequence are 4, 3x, 3xPage
+ 5
16, 15x + 16 and 27x + 80.
ANSWER: 5, 6, 8, 8 and Iteration
10-54,Recursion
ANSWER: 3, 2x, 8x – 9, 26x – 36, 80x – 117
20. a 1 = 4, a 2 = 3x, an = a n – 1 + 4a n – 2
22. a 1 = 2, a 2 = x + 3, a n = a n – 1 + 6a n – 2
SOLUTION: SOLUTION: The first five terms of the sequence are 4, 3x, 3x +
16, 15x + 16 and 27x + 80.
The first five terms of the sequence are 2, x + 3, x
+15, 7x + 33 and 13x + 123.
ANSWER: 4, 3x, 3x + 16, 15x + 16, 27x + 80
ANSWER: 2, x + 3, x + 15, 7x + 33, 13x + 123
21. a 1 = 3, a 2 = 2x, an = 4a n – 1 – 3a n – 2
23. a 1 = 1, a 2 = x, an = 3a n – 1 + 6a n – 2
SOLUTION: SOLUTION: The first five terms of the sequence are 3, 2x, 8x – 9,
26x – 36 and 80x –117.
The first five terms of the sequence are 1, x, 3x +6,
15x + 18 and 63x + 90.
ANSWER: 3, 2x, 8x – 9, 26x – 36, 80x – 117
ANSWER: 1, x, 3x + 6, 15x + 18, 63x + 90
22. a 1 = 2, a 2 = x + 3, a n = a n – 1 + 6a n – 2
Write a recursive formula for each sequence.
SOLUTION: 24. 16, 10, 7, 5.5, 4.75, …
SOLUTION: The Manual
first five
termsby
ofCognero
the sequence
eSolutions
- Powered
+15, 7x + 33 and 13x + 123.
are 2, x + 3, x
Page 6
ANSWER: 1, x, 3x + 6, 15x + 18, 63x + 90
10-5 Recursion and Iteration
ANSWER: a n + 1 = 0.25a n + 4; a 1 = 32
Write a recursive formula for each sequence.
26. 4, 15, 224, 50,175, …
24. 16, 10, 7, 5.5, 4.75, …
SOLUTION: SOLUTION: ANSWER: 2
a n + 1 = (a n) – 1; a 1 = 4
ANSWER: a n + 1 = 0.5a n + 2; a 1 = 16
27. 1, 2, 9, 730, …
25. 32, 12, 7, 5.75, …
SOLUTION: SOLUTION: ANSWER: 3
ANSWER: a n + 1 = 0.25a n + 4; a 1 = 32
a n + 1 = (a n) + 1; a 1 = 1
28. 9, 33, 129, 513, …
26. 4, 15, 224, 50,175, …
SOLUTION: SOLUTION: ANSWER: a n + 1 = 4a n – 3; a 1 = 9
ANSWER: eSolutions Manual - Powered by Cognero
2
a n + 1 = (a n) – 1; a 1 = 4
29. 480, 128, 40, 18, …
Page 7
ANSWER: 3
+ 1;Iteration
a1 = 1
n + 1 = (a n)and
10-5aRecursion
ANSWER: a n + 1 = 0.25a n + 8; a 1 = 480
28. 9, 33, 129, 513, …
30. 393, 132, 45, 16, …
SOLUTION: SOLUTION: ANSWER: a n + 1 = 4a n – 3; a 1 = 9
ANSWER: 29. 480, 128, 40, 18, …
; a 1 = 393
SOLUTION: 31. 68, 104, 176, 320, …
SOLUTION: ANSWER: a n + 1 = 0.25a n + 8; a 1 = 480
ANSWER: a n + 1 = 2a n – 32; a 1 = 84
SOLUTION: 30. 393, 132, 45, 16, …
32. FINANCIAL LITERACY Mr. Edwards and his
company deposit $20,000 into his retirement account
at the end of each year. The account earns 8%
interest before each deposit.
a. Write a recursive formula for the balance in the
account at the end of each year.
b. Determine how much is in the account at the end
of each of the first 8 years.
ANSWER: eSolutions Manual - Powered
by Cognero
; a = 393
1
SOLUTION: a. Let a n represent the balance on the account inPage
the 8
nth year. The initial balance a 0 is $0.
The recursive formula is a = 1.08a
+ 20000.
a 6 = $146,718.58,
b. Determine how much is in the account at the end
of each of the first 8 years.
10-5 Recursion and Iteration
SOLUTION: a. Let a n represent the balance on the account in the
nth year. The initial balance a 0 is $0.
The recursive formula is a n = 1.08a n – 1 + 20000.
a 7 = $178,456.07,
a 8 = $212,732.56
Find the first three iterates of each function for
the given initial value.
33. f (x) = 12x + 8, x 0 = 4
b.
SOLUTION: The first three iterates are 56, 680 and 8168.
ANSWER: 56, 680, 8168
34. f (x) = –9x + 1, x 0 = –6
SOLUTION: ANSWER: a. a n = 1.08a n – 1 + 20000
b.
a 1 = $20,000,
a 2 = $41,600,
a 3 = $64,928,
The first three iterates are 55, –494 and 4447.
a 4 = $90,122.24,
a 5 = $117,332.02,
a 6 = $146,718.58,
ANSWER: 55, –494, 4447
a 7 = $178,456.07,
a 8 = $212,732.56
35. f (x) = –6x + 3, x0 = 8
Find the first three iterates of each function for
the given initial value.
SOLUTION: 33. f (x) = 12x + 8, x 0 = 4
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SOLUTION: Page 9
ANSWER: –494, 4447
10-555,
Recursion
and Iteration
ANSWER: –29, –229, –1829
35. f (x) = –6x + 3, x0 = 8
2
37. f (x) = –3x + 9, x 0 = 2
SOLUTION: SOLUTION: The first three iterates are –45, 273 and –1635.
The first three iterates are –3, –18 and –963.
ANSWER: –45, 273, –1635
ANSWER: –3, –18, –963
36. f (x) = 8x + 3, x0 = –4
SOLUTION: 2
38. f (x) = 4x + 5, x 0 = –2
SOLUTION: The first three iterates are –29, –229 and –1829.
The first three iterates are 21, 1769 and 12,517,449.
ANSWER: –29, –229, –1829
2
37. f (x) = –3x + 9, x 0 = 2
SOLUTION: ANSWER: 21, 1769, 12,517,449
2
39. f (x) = 2x – 5x + 1, x0 = 6
SOLUTION: eSolutions Manual - Powered by Cognero
Page 10
ANSWER: 1769, 12,517,449
10-521,
Recursion
and Iteration
ANSWER: –2, 3, 6.75
2
39. f (x) = 2x – 5x + 1, x0 = 6
2
41. f (x) = x + 2x + 3,
SOLUTION: SOLUTION: The first three iterates are 43, 3484 and 24,259,093.
The first three iterates are 4.25, 29.5625 and
936.06640625.
ANSWER: 43, 3484, 24,259,093
2
ANSWER: 4.25, 29.5625, 936.0664
40. f (x) = –0.25x + x + 6, x0 = 8
2
42. f (x) = 2x + x + 1,
SOLUTION: SOLUTION: The first three iterates are –2, 3 and 6.75.
ANSWER: –2, 3, 6.75
2
41. f (x) = x + 2x + 3,
SOLUTION: The first three iterates are 1, 4 and 37.
ANSWER: 1, 4, 37
43. FRACTALS Consider the figures at the right. The
number of blue triangles increases according to a
specific pattern.
eSolutions Manual - Powered by Cognero
Page 11
a. Write a recursive formula for the number of blue
a. a n = 3a n – 1; a 1 =1
ANSWER: 4, 37
10-51,Recursion
and Iteration
b. 243
43. FRACTALS Consider the figures at the right. The
number of blue triangles increases according to a
specific pattern.
44. FINANCIAL LITERACY Miguel’s monthly car
payment is $234.85. The recursive formula b n =
1.005b n – 1 – 234.85 describes the balance left on the
loan after n payments. Find the balance of the
$10,000 loan after each of the first eight payments.
a. Write a recursive formula for the number of blue
triangles in the sequence of figures.
SOLUTION: Given b 0 = 10,000.
b. How many blue triangles will be in the sixth
figure?
SOLUTION: a. The number triangles in the figures are 1, 3 and 9.
Therefore, the recursive formula is
.
b.
ANSWER: a. a n = 3a n – 1; a 1 =1
b. 243
44. FINANCIAL LITERACY Miguel’s monthly car
payment is $234.85. The recursive formula b n =
1.005b n – 1 – 234.85 describes the balance left on the
loan after n payments. Find the balance of the
$10,000 loan after each of the first eight payments.
eSolutions
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SOLUTION: ANSWER: $9815.15, $9629.38, $9442.67, $9255.04, $9066.46,
$8876.94, $8686.48, $8495.06
45. CONSERVATION Suppose a lake is populated
with 10,000 fish. A year later, 80% of the fish have
died or been caught, and the lake is replenishedPage
with12
10,000 new fish. If the pattern continues, will the
lake eventually run out of fish? If not, will the
ANSWER: $9815.15, $9629.38, $9442.67, $9255.04, $9066.46,
$8686.48,
$8495.06
10-5$8876.94,
Recursion
and Iteration
45. CONSERVATION Suppose a lake is populated
with 10,000 fish. A year later, 80% of the fish have
died or been caught, and the lake is replenished with
10,000 new fish. If the pattern continues, will the
lake eventually run out of fish? If not, will the
population of the lake converge to any particular
value? Explain.
ANSWER: No; the population of fish will reach 12,500. Each
year, 20% of 12,500 or 2500 fish, plus 10,000
additional fish, yields 12,500 fish.
46. GEOMETRY Consider the pattern.
a. Write a sequence of the total number of triangles
in the first six figures.
SOLUTION: No; the population of fish will reach 12,500. Each
year, 20% of 12,500 or 2500 fish, plus 10,000
additional fish, yields 12,500 fish.
c. How many triangles will be in the tenth figure?
ANSWER: No; the population of fish will reach 12,500. Each
year, 20% of 12,500 or 2500 fish, plus 10,000
additional fish, yields 12,500 fish.
b. Write a recursive formula for the number of
triangles.
SOLUTION: a. a 1 = 1, a 2 = 4, a 3 = 10,
46. GEOMETRY Consider the pattern.
Therefore:
a. Write a sequence of the total number of triangles
in the first six figures.
b. Write a recursive formula for the number of
triangles.
b. The recursive formula is a n = a n – 1 + 3(n – 1)
c.
c. How many triangles will be in the tenth figure?
SOLUTION: a. a 1 = 1, a 2 = 4, a 3 = 10,
The number of triangles in the tenth figure is 136.
Therefore:
ANSWER: a. 1, 4, 10, 19, 31, 46
b. a n = a n – 1 + 3(n – 1)
b. The recursive formula is a n = a n – 1 + 3(n – 1)
eSolutions
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c.
c. 136
Page
47. SPREADSHEETS Consider the sequence with
x0 13
=
20,000 and f (x) = 0.3x + 5000.
b. a n = a n – 1 + 3(n – 1)
136
10-5c.Recursion
and Iteration
47. SPREADSHEETS Consider the sequence with x0 =
20,000 and f (x) = 0.3x + 5000.
a. Enter x0 in cell A1 of your spreadsheet. Enter “=
(0.3)*(A1) + 5000” in cell A2. What answer does it
provide?
b. Copy cell A2, highlight cells A3 through A70, and
paste. What do you notice about the sequence?
c. How do spreadsheets help analyze recursive
sequences?
SOLUTION: a. 11,000
b. It converges to 7142.857.
c. Sample answer: They make it easier to analyze
recursive sequences because they can produce the
first 100 terms instantaneously; it would take a long
time to calculate the terms by hand.
c. Sample answer: They make it easier to analyze
recursive sequences because they can produce the
first 100 terms instantaneously; it would take a long
time to calculate the terms by hand.
48. VIDEO GAMES The final monster in Helena’s
video game has 100 health points. During the final
battle, the monster regains 10% of its health points
after every 10 seconds. If Helena can inflict damage
to the monster that takes away 10 health points every
10 seconds without getting hurt herself, will she ever
kill the monster? If so, when?
SOLUTION: Let’s assume that the monster dies when its health
points reach 0 and it starts the battle with all 100 of
its health points. After the first 10-seconds, the
monster would have (100 - 10) + 0.1(100 - 10) health
points, which is 1.1(100 - 10) or 99. After 20seconds, the monster would have 1.1(99 - 10) or 97.9
health points. Organize this information into a table
and complete the table through 50 seconds. Then use
the table to write a recursive formula to represent the
number of health points the monster has after n 10second time periods.
Time
(10-second time
periods)
1
2
3
ANSWER: a. 11,000
4
b. It converges to 7142.857.
5
c. Sample answer: They make it easier to analyze
recursive sequences because they can produce the
first 100 terms instantaneously; it would take a long
time to calculate the terms by hand.
M
n
M
25
26
48. VIDEO GAMES The final monster in Helena’s
video game has 100 health points. During the final
battle, the monster regains 10% of its health points
after every 10 seconds. If Helena can inflict damage
to the monster that takes away 10 health points every
10 seconds without getting hurt herself, will she ever
kill the monster? If so, when?
SOLUTION: Let’s assume that the monster dies when its health
points reach 0 and it starts the battle with all 100 of
its health points. After the first 10-seconds, the
monster
would
haveby(100
- 10) + 0.1(100 - 10) health
eSolutions
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points, which is 1.1(100 - 10) or 99. After 20seconds, the monster would have 1.1(99 - 10) or 97.9
health points. Organize this information into a table
Monster’s Health
Points
1.1(100 - 10) or 99
1.1(99 - 10) or 97.9
1.1(97.9 - 10) or 96.69
1.1(96.69 - 10) or
95.359
1.1(95.359 - 10) or
93.849
M
a n = 1.1(a n - 1 - 10)
M
≈ 1.65
≈-9.18
Between n = 25 and n = 26, the monster’s health
points drop to zero. Since n represents the number of
10-second time periods, the monster’s health points
drop to zero and the monster is defeated between 25
(10) and 26(10) seconds or between 250 and 260
seconds.
ANSWER: Yes; between 250 and 260 seconds
49. CCSS CRITIQUE Marcus and Armando arePage 14
finding the first three iterates of f (x) = 5x – 3 for an
initial value of x0 = 4. Is either of them correct?
ANSWER: betweenand
250Iteration
and 260 seconds
10-5Yes;
Recursion
49. CCSS CRITIQUE Marcus and Armando are
finding the first three iterates of f (x) = 5x – 3 for an
initial value of x0 = 4. Is either of them correct?
Explain.
Armando; Marcus included x0 with the iterates and
only showed the first 2 iterates.
50. CHALLENGE Find a recursive formula for 5, 23,
98, 401, … .
SOLUTION: ANSWER: a n + 1 = 4a n + 3n; a1 =5
51. REASONING Is the statement “If the first three
terms of a sequence are identical, then the
sequence is not recursive” sometimes, always, or
never true? Explain your reasoning.
SOLUTION: Sample answer: Sometimes; the recursive formula
could involve the first three terms. For example, 2, 2,
2, 8, 20,… . is recursive with a n + 3 = a n + a n + 1 +
SOLUTION: Armando; Marcus included x0 with the iterates and
2a n + 2.
only showed the first 2 iterates.
ANSWER: Sample answer: Sometimes; the recursive formula
could involve the first three terms. For example, 2, 2,
2, 8, 20, . . . is recursive with a n + 3 = a n + a n + 1 +
ANSWER: Armando; Marcus included x0 with the iterates and
only showed the first 2 iterates.
50. CHALLENGE Find a recursive formula for 5, 23,
98, 401, … .
2a n + 2.
52. OPEN ENDED Write a function for which the first
three iterates are 9, 19, and 39.
SOLUTION: Sample answer: f (x) = 2x + 1, x0 = 4
SOLUTION: ANSWER: Sample answer: f (x) = 2x + 1, x0 = 4
eSolutions Manual - Powered by Cognero
Page 15
53. WRITING IN MATH Why is it useful to represent
a sequence with an explicit or recursive formula?
ANSWER: answer:
(x) = 2x + 1, x0 = 4
10-5Sample
Recursion
andf Iteration
53. WRITING IN MATH Why is it useful to represent
a sequence with an explicit or recursive formula?
SOLUTION: Sample answer: In a recursive sequence, each term
is determined by one or more of the previous terms.
A recursive formula is used to produce the terms of
the recursive sequence.
ANSWER: Sample answer: In a recursive sequence, each term
is determined by one or more of the previous terms.
A recursive formula is used to produce the terms of
the recursive sequence.
54. GEOMETRY In the figure shown, a + b + c = ?
ANSWER: C
55. EXTENDED RESPONSE Bill launches a model
rocket from ground level. The rocket’s height h in
2
meters is given by the equation h = –4.9t + 56t,
where t is the time in seconds after the launch.
a. What is the maximum height the rocket will reach?
b. How long after it is launched will the rocket reach
its maximum height? Round to the nearest tenth of a
second.
c. How long after it is launched will the rocket land?
Round to the nearest tenth of a second.
SOLUTION: a. Substitute 0 for h and find the vertex of the
equation.
The vertex of a quadratic equation is
.
A
B
C
D
SOLUTION: The sum of the exterior angles of a polygon is 360°.
Therefore, option C is the correct answer.
The vertex of the equation is (5.7, 160).
Therefore, the rocket will reach the maximum height
of 160 m.
b. The vertex of the equation is (5.7, 160).
Therefore, it will take 5.7 s to reach the maximum
height.
c. The rocket will land in 5.7 × 2 or 11.4 s after it is
launched.
ANSWER: C
ANSWER: a. 160 m
b. 5.7 s
55. EXTENDED RESPONSE Bill launches a model
rocket from ground level. The rocket’s height h in
2
meters is given by the equation h = –4.9t + 56t,
where t is the time in seconds after the launch.
eSolutions Manual - Powered by Cognero
a. What is the maximum height the rocket will reach?
c. 11.4 s
of y 16
=
56. Which of the following is true about the graphs Page
2
2
3(x – 4) + 5 and y = 3(x + 4) + 5?
b. 5.7 s
ANSWER: G
10-5c.Recursion
and Iteration
11.4 s
56. Which of the following is true about the graphs of y =
2
2
3(x – 4) + 5 and y = 3(x + 4) + 5?
57. Which factors could represent the length times the
width?
F Their vertices are maximums.
G The graphs have the same shape with different
vertices.
A (4x – 5y)(4x – 5y)
H The graphs have different shapes with different
vertices.
B (4x + 5y)(4x – 5y)
J One graph has a vertex that is a maximum, while
the other graph has a vertex that is a minimum.
C (4x2 – 5y)(4x2 + 5y)
SOLUTION: The graphs have the same shape with different
vertices.
Therefore, option G is the correct answer.
D (4x + 5y)(4x + 5y)
2
2
SOLUTION: Factor the area of the rectangle.
ANSWER: G
57. Which factors could represent the length times the
width?
Option C is the correct answer.
ANSWER: C
Write each repeating decimal as a fraction.
A (4x – 5y)(4x – 5y)
58. B (4x + 5y)(4x – 5y)
C (4x2 – 5y)(4x2 + 5y)
SOLUTION: 2
2
D (4x + 5y)(4x + 5y)
SOLUTION: Factor the area of the rectangle.
Find the value of r.
eSolutions
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Option
C is- Powered
the correct
answer.
Page 17
ANSWER: ANSWER: 10-5CRecursion and Iteration
Write each repeating decimal as a fraction.
59. 58. SOLUTION: The number
SOLUTION: can be written as .
Find the value of r.
Find the value of r.
Therefore:
ANSWER: ANSWER: 59. SOLUTION: The number
can be written as .
60. SOLUTION: Find the value of r.
eSolutions Manual - Powered by Cognero
The number
can be written as .
Find the value of r.
Page 18
a. Write the first five terms of a sequence describing
his training schedule.
ANSWER: 10-5 Recursion and Iteration
b. When will he exceed 26 miles in one run?
c. When will he have run 100 total miles?
60. SOLUTION: SOLUTION: a. This forms a geometric sequence.
The number
can be written as .
Given a 1 = 2, and r = 1.5
Find the value of r.
b. Given a n = 26
Find n.
He will exceed 26 miles in one run on the eighth
session.
Therefore:
c. Given S n = 100
Find n.
ANSWER: 61. SPORTS Adrahan is training for a marathon, about
26 miles. He begins by running 2 miles. Then, when
he runs every other day, he runs one and a half times
the distance he ran the time before.
a. Write the first five terms of a sequence describing
his training schedule.
b. When will he exceed 26 miles in one run?
SOLUTION: He will have run 100 total miles during the ninth
session.
c. When
will
he have
100
eSolutions
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Cognero
total miles?
ANSWER: a. 2, 3, 4.5, 6.75, 10.125
b. the eighth session
c. during the ninth session
Page 19
ANSWER: a. 2, 3, 4.5, 6.75, 10.125
10-5 Recursion and Iteration
b. the eighth session
c. during the ninth session
ANSWER: 2
y + 7y + 12
65. (x – 2)(x + 6)
SOLUTION: State whether the events are independent or
dependent.
62. tossing a penny and rolling a number cube
ANSWER: SOLUTION: They are independent events.
ANSWER: independent
2
x + 4x – 12
66. (a – 8)(a + 5)
SOLUTION: 63. choosing first and second place in an academic
competition
SOLUTION: They are dependent events.
ANSWER: 2
a – 3a – 40
ANSWER: dependent
67. (4h + 5)(h + 7)
SOLUTION: Find each product.
64. (y + 4)(y + 3)
SOLUTION: ANSWER: 2
4h + 33h + 35
68. (9p – 1)(3p – 2)
ANSWER: 2
SOLUTION: y + 7y + 12
65. (x – 2)(x + 6)
ANSWER: SOLUTION: 2
27p – 21p + 2
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eSolutions
ANSWER: 2
69. (2g + 7)(5g – 8)
SOLUTION: Page 20
ANSWER: 2
2 Iteration
– 21p +and
10-527p
Recursion
69. (2g + 7)(5g – 8)
SOLUTION: ANSWER: 2
10g + 19g – 56
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Page 21
ANSWER: 7
6
5 2
4 3
3 4
2 5
g + 7g h + 21g h + 35g h + 35g h + 21g h +
6
7gh + h
10-6 The Binomial Theorem
7
6
Expand each binomial.
3. (x – 4)
1. (c + d)
5
SOLUTION: Replace n with 6 in the Binomial Theorem. SOLUTION: Replace n with 5 in the Binomial Theorem.
ANSWER: 6
5
4
3
2
x – 24x + 240x – 1280x + 3840x – 6144x +
4096
ANSWER: 5
4
3 2
2 3
4
c + 5c d + 10c d + 10c d + 5cd + d
5
2. (g + h)
5
4. (2y – z)
SOLUTION: Replace n with 5 in the Binomial Theorem.
7
SOLUTION: Replace n with 7 in the Binomial Theorem. ANSWER: 5
4
3 2
2 3
4
32y – 80y z + 80y z – 40y z + 10yz – z
5
5
5. (x + 3)
ANSWER: 7
6
5 2
4 3
3 4
2 5
g + 7g h + 21g h + 35g h + 35g h + 21g h +
6
7gh + h
7
SOLUTION: Replace n with 5 in the Binomial Theorem.
6
3. (x – 4)
SOLUTION: Replace n with 6 in the Binomial Theorem. eSolutions Manual - Powered by Cognero
Page 1
ANSWER: ANSWER: 5
4
3 2
2 3
4
80y z + 80y
z – 40y z + 10yz – z
10-632y
The –Binomial
Theorem
5
4
3
2 2
3
y – 16y z + 96y z – 256yz + 256z
4
5
5. (x + 3)
SOLUTION: Replace n with 5 in the Binomial Theorem.
7. GENETICS If a woman is equally as likely to have
a baby boy or a baby girl, use binomial expansion to
determine the probability that 5 of her 6 children are
girls. Do not consider identical twins.
SOLUTION: 6
Find the probability by expanding (g + b) .
By adding the coefficients of the polynomial, we
determine that there are 64 combinations of girls and
boys.
5
6g b represents the number of combinations with 5
girls and 1 boy.
ANSWER: 5
4
3
2
x + 15x + 90x + 270x + 405x + 243
6. (y – 4z)
4
Therefore, the probability that 5 of her 6 children are
girls is 0.09375.
SOLUTION: Replace n with 4 in the Binomial Theorem.
ANSWER: or 0.09375
Find the indicated term of each expression.
8. fourth term of (b + c)
9
SOLUTION: Use the Binomial Theorem to write the expansion in
sigma notation.
ANSWER: 4
3
2 2
3
y – 16y z + 96y z – 256yz + 256z
4
7. GENETICS If a woman is equally as likely to have
a baby boy or a baby girl, use binomial expansion to
determine the probability that 5 of her 6 children are
girls. Do not consider identical twins.
For the fourth term k = 3.
SOLUTION: 6
Find the probability by expanding (g + b) .
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ANSWER: Page 2
ANSWER: ANSWER: or 0.09375
10-6 The Binomial Theorem
4 4
5670x y
Find the indicated term of each expression.
8. fourth term of (b + c)
9
SOLUTION: Use the Binomial Theorem to write the expansion in
sigma notation.
10. third term of (a – 4b)
6
SOLUTION: Use the Binomial Theorem to write the expansion in
sigma notation.
For the third term k = 2.
For the fourth term k = 3.
ANSWER: 4 2
240a b
ANSWER: 6 3
84b c
11. sixth term of (2c – 3d)
8
9. fifth term of (x + 3y)
8
SOLUTION: Use the Binomial Theorem to write the expansion in
sigma notation.
SOLUTION: Use the Binomial Theorem to write the expansion in
sigma notation.
For the sixth term k = 5.
For the fifth term k = 4.
ANSWER: 3 5
–108,864c d
ANSWER: 4 4
5670x y
12. last term of (5x + y)
10. third term of (a – 4b)
6
SOLUTION: Use the Binomial Theorem to write the expansion in
sigma notation.
eSolutions Manual - Powered by Cognero
5
SOLUTION: Use the Binomial Theorem to write the expansion in
sigma notation.
Page 3
ANSWER: ANSWER: 3 5
d
10-6–108,864c
The Binomial
Theorem
243a
5
12. last term of (5x + y)
5
SOLUTION: Use the Binomial Theorem to write the expansion in
sigma notation.
14. CCSS MODELING The color of a particular
flower is determined by the combination of two
genes, also called alleles. If the flower has two red
alleles r, the flower is red. If the flower has two
white alleles w, the flower is white. If the flower has
one allele of each color, the flower will be pink. In a
lab, two pink flowers are mated and eventually
produce 1000 offspring. How many of the 1000
offspring will be pink?
For the last term k = 5.
ANSWER: y
5
13. first term of (3a + 8b)
5
SOLUTION: Here n = 1000.
SOLUTION: Use the Binomial Theorem to write the expansion in
sigma notation.
Find the term for which the powers or r and w are
equal.
For the first term k = 0.
Therefore, there will be 500 pink flowers in 1000
offspring.
ANSWER: 243a
5
14. CCSS MODELING The color of a particular
flower is determined by the combination of two
genes, also called alleles. If the flower has two red
alleles r, the flower is red. If the flower has two
white alleles w, the flower is white. If the flower has
one allele of each color, the flower will be pink. In a
lab, two pink flowers are mated and eventually
produce
1000
offspring.
How many of the 1000
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offspring will be pink?
ANSWER: 500
Expand each binomial.
15. (a – b)
6
SOLUTION: Replace n with 6 in the Binomial Theorem.
Page 4
ANSWER: 7
6
5 2
4 3
3 4
2 5
c – 7c d + 21c d – 35c d + 35c d – 21c d +
ANSWER: 10-6500
The Binomial Theorem
6
7cd – d
7
6
Expand each binomial.
17. (x + 6)
15. (a – b)
6
SOLUTION: Replace n with 6 in the Binomial Theorem.
SOLUTION: Replace n with 6 in the Binomial Theorem.
ANSWER: 6
5
4 2
3 3
2 4
5
a – 6a b + 15a b – 20a b + 15a b – 6ab + b
6
ANSWER: 6
5
4
3
2
x + 36x + 540x + 4320x + 19,440x + 46,656x +
46,656
16. (c – d)
7
7
18. (y – 5)
SOLUTION: Replace n with 7 in the Binomial Theorem.
SOLUTION: Replace n with 7 in the Binomial Theorem.
ANSWER: 7
6
5 2
4 3
3 4
2 5
c – 7c d + 21c d – 35c d + 35c d – 21c d +
6
7cd – d
7
ANSWER: 7
6
5
4
3
2
y – 35y + 525y – 4375y + 21,875y – 65,625y +
109,375y – 78,125
6
17. (x + 6)
19. (2a + 4b)
SOLUTION: Replace n with 6 in the Binomial Theorem.
SOLUTION: Replace n with 4 in the Binomial Theorem.
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4
Page 5
ANSWER: ANSWER: 7
6
5
4
3
5
2
y – 35y + 525y – 4375y + 21,875y – 65,625y +
– 78,125
10-6109,375y
The Binomial
Theorem
4
3 2
2 3
4
243a – 1620a b + 4320a b – 5760a b + 3840ab
– 1024b
5
19. (2a + 4b)
4
21. COMMITTEES If an equal number of men and
women applied to be on a community planning
committee and the committee needs a total of 10
people, find the probability that 7 of the members will
be women. Assume that committee members will be
chosen randomly.
SOLUTION: Replace n with 4 in the Binomial Theorem.
SOLUTION: 10
Find the probability by expanding (m + w) .
By adding the coefficients of the polynomial, we
determine that there are 1024 combinations of men
and women.
ANSWER: 4
3
2 2
3
16a + 128a b + 384a b + 512ab + 256b
4
3 7
20. (3a – 4b)
120m w represents the number of combinations with
7 women and 3 men.
5
Therefore, the probability that 7 of the members are
SOLUTION: Replace n with 5 in the Binomial Theorem.
women is
.
ANSWER: 22. BASEBALL If a pitcher is just as likely to throw a
ball as a strike, find the probability that 11 of his first
12 pitches are balls.
ANSWER: 5
4
3 2
2 3
4
243a – 1620a b + 4320a b – 5760a b + 3840ab
– 1024b
5
SOLUTION: 12
Find the probability by expanding (b + p ) .
21. COMMITTEES If an equal number of men and
women applied to be on a community planning
committee and the committee needs a total of 10
people, find the probability that 7 of the members will
be women. Assume that committee members will be
chosen randomly.
By adding the coefficients of the polynomial, we
determine that there are 4096 combinations of balls
and pitches.
11
SOLUTION: 10
Find the probability by expanding (m + w) .
12b p represents the number of combinations with
11 balls.
Therefore, the probability that 11 of his first 12
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Page 6
pitches are balls is
.
ANSWER: ANSWER: 10-6 The Binomial Theorem
22. BASEBALL If a pitcher is just as likely to throw a
ball as a strike, find the probability that 11 of his first
12 pitches are balls.
Find the indicated term of each expression.
23. third term of (x + 2z)
7
SOLUTION: SOLUTION: Use the Binomial Theorem to write the expansion in
sigma notation.
12
Find the probability by expanding (b + p ) .
By adding the coefficients of the polynomial, we
determine that there are 4096 combinations of balls
and pitches.
For the third term k = 2.
11
12b p represents the number of combinations with
11 balls.
Therefore, the probability that 11 of his first 12
pitches are balls is
.
ANSWER: 5 2
84x z
ANSWER: 24. fourth term of (y – 3x)
6
Find the indicated term of each expression.
23. third term of (x + 2z)
SOLUTION: Use the Binomial Theorem to write the expansion in
sigma notation.
7
SOLUTION: Use the Binomial Theorem to write the expansion in
sigma notation.
For the fourth term k = 3.
For the third term k = 2.
ANSWER: 3 3
–540y x
25. seventh term of (2a – 2b)
8
ANSWER: 5 2
84x z
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SOLUTION: Use the Binomial Theorem to write the expansion in
Page 7
sigma notation.
6
ANSWER: ANSWER: 3 3
5
x
10-6–540y
The Binomial
Theorem
75,000xy
25. seventh term of (2a – 2b)
8
9
27. fifth term of (x – 4)
SOLUTION: Use the Binomial Theorem to write the expansion in
sigma notation.
SOLUTION: Use the Binomial Theorem to write the expansion in
sigma notation.
For the seventh term k = 6.
For the fifth term k = 4.
ANSWER: ANSWER: 2 6
7168a b
32,256x
5
26. sixth term of (4x + 5y)
6
28. fourth term of (c + 6)
8
SOLUTION: Use the Binomial Theorem to write the expansion in
sigma notation.
SOLUTION: Use the Binomial Theorem to write the expansion in
sigma notation.
For the sixth term k = 5.
For the fourth term k = 3.
ANSWER: ANSWER: 5
75,000xy
12,096c
5
9
27. fifth term of (x – 4)
Expand each binomial.
SOLUTION: Use the Binomial Theorem to write the expansion in
sigma notation.
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29. SOLUTION: Replace n with 5 in the Binomial Theorem.
Page 8
ANSWER: ANSWER: 5
10-612,096c
The Binomial Theorem
Expand each binomial.
30. 29. SOLUTION: Replace n with 4 in the Binomial Theorem.
SOLUTION: Replace n with 5 in the Binomial Theorem.
ANSWER: ANSWER: 31. SOLUTION: Replace n with 5 in the Binomial Theorem.
30. SOLUTION: Replace n with 4 in the Binomial Theorem.
ANSWER: ANSWER: 32. eSolutions Manual - Powered by Cognero
SOLUTION: Replace n with 5 in the Binomial Theorem.
31. Page 9
SOLUTION: a. Substitute n = 10, k = 9, p = 0.7 and q = 0.3 in
ANSWER: .
10-6 The Binomial Theorem
32. SOLUTION: Replace n with 5 in the Binomial Theorem.
b. Substitute n = 10, k = 8, p = 0.6 and q = 0.4 in
.
ANSWER: c. Substitute n = 5, k = 2, p = 0.3 and q = 0.7 in
.
33. CCSS SENSE-MAKING In
, let
p represent the likelihood of a success and q
represent the likelihood of a failure.
a. If a place-kicker makes 70% of his kicks within 40
yards, find the likelihood that he makes 9 of his next
10 attempts from within 40 yards.
ANSWER: b. If a quarterback completes 60% of his passes,
find the likelihood that he completes 8 of his next 10
attempts.
b. 0.121
c. 0.309
c. If a team converts 30% of their two-point
conversions, find the likelihood that they convert 2 of
their next 5 conversions.
a. 0.121
34. CHALLENGE Find the sixth term of the expansion
of
SOLUTION: a. Substitute n = 10, k = 9, p = 0.7 and q = 0.3 in
.
. Explain your reasoning.
SOLUTION: Use the Binomial Theorem to write the expansion in
sigma notation.
For the sixth term k = 5.
eSolutions Manual - Powered by Cognero
b. Substitute n = 10, k = 8, p = 0.6 and q = 0.4 in
.
Page 10
b. 0.121
Sample answer: While they have the same terms, the
signs for (x + y) will all be positive, while the signs
n
n
. 0.309
10-6cThe
Binomial Theorem
for (x – y) will alternate.
34. CHALLENGE Find the sixth term of the expansion
of
. Explain your reasoning.
36. REASONING Determine whether the following
statement is true or false. Explain your reasoning.
SOLUTION: Use the Binomial Theorem to write the expansion in
sigma notation.
The eighth and twelfth terms of (x + y)
same coefficients.
20
have the
SOLUTION: th
Sample answer: False; a binomial to the 20 power
will have 21 terms. The eleventh term will be in the
middle and the rest of the terms will be symmetric.
For the sixth term k = 5.
The tenth term corresponds with the twelfth term.
ANSWER: th
Sample answer: False; a binomial to the 20 power
will have 21 terms. The eleventh term will be in the
ANSWER: middle and the rest of the terms will be symmetric.
The tenth term corresponds with the twelfth term.
35. REASONING Explain how the terms of (x + y)
n n
and (x – y) are the same and how they are
different.
37. OPEN ENDED Write a power of a binomial for
4
which the second term of the expansion is 6x y.
SOLUTION: SOLUTION: Sample answer: While they have the same terms, the
n
signs for (x + y) will all be positive, while the signs
n
Sample answer:
for (x – y) will alternate.
ANSWER: ANSWER: Sample answer:
Sample answer: While they have the same terms, the
n
signs for (x + y) will all be positive, while the signs
38. WRITING IN MATH Explain how to write out the
terms of Pascal’s triangle.
n
for (x – y) will alternate.
36. REASONING Determine whether the following
statement is true or false. Explain your reasoning.
The eighth and twelfth terms of (x + y)
same coefficients.
20
have the
SOLUTION: Sample answer: The first row is a 1. The second row
is two 1s. Each new row begins and ends with 1.
Each coefficient is the sum of the two coefficients
above it in the previous row.
SOLUTION: th
Sample answer: False; a binomial to the 20 power
will have 21 terms. The eleventh term will be in the
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middle and the rest of the terms will be symmetric.
ANSWER: Sample answer: The first row is a 1. The second row
Page 11
is two 1s. Each new row begins and ends with 1.
Each coefficient is the sum of the two coefficients
above it in the previous row.
ANSWER: Sample answer:
10-6 The Binomial Theorem
38. WRITING IN MATH Explain how to write out the
terms of Pascal’s triangle.
SOLUTION: Sample answer: The first row is a 1. The second row
is two 1s. Each new row begins and ends with 1.
Each coefficient is the sum of the two coefficients
above it in the previous row.
Sample answer: The first row is a 1. The second row
is two 1s. Each new row begins and ends with 1.
Each coefficient is the sum of the two coefficients
above it in the previous row.
39. PROBABILITY A desk drawer contains 7
sharpened red pencils, 5 sharpened yellow pencils, 3
unsharpened red pencils, and 5 unsharpened yellow
pencils. If a pencil is taken from the drawer at
random, what is the probability that it is yellow, given
that it is one of the sharpened pencils?
A
ANSWER: Sample answer: The first row is a 1. The second row
is two 1s. Each new row begins and ends with 1.
Each coefficient is the sum of the two coefficients
above it in the previous row.
B
39. PROBABILITY A desk drawer contains 7
sharpened red pencils, 5 sharpened yellow pencils, 3
unsharpened red pencils, and 5 unsharpened yellow
pencils. If a pencil is taken from the drawer at
random, what is the probability that it is yellow, given
that it is one of the sharpened pencils?
A
C
D
SOLUTION: Favorable outcomes: {5 sharpened yellow pencils}
Possible outcomes: {7 sharpened red pencils, 5
sharpened yellow pencils}
Probability B
A is the correct option.
C
D
SOLUTION: Favorable outcomes: {5 sharpened yellow pencils}
Possible outcomes: {7 sharpened red pencils, 5
sharpened yellow pencils}
Probability A is the correct option.
ANSWER: A
40. GRIDDED RESPONSE Two people are 17.5 miles
apart. They begin to walk toward each other along a
straight line at the same time. One walks at the rate
of 4 miles per hour, and the other walks at the rate of
3 miles per hour. In how many hours will they meet?
SOLUTION: Let the distance traveled by one person be x.
So, the distance traveled by the other person be 17.5
– x.
The equation that represents this situation is
ANSWER: A
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40. GRIDDED RESPONSE Two people are 17.5 miles
apart. They begin to walk toward each other along a
.
Page 12
ANSWER: ANSWER: 10-6AThe Binomial Theorem
2.5
40. GRIDDED RESPONSE Two people are 17.5 miles
apart. They begin to walk toward each other along a
straight line at the same time. One walks at the rate
of 4 miles per hour, and the other walks at the rate of
3 miles per hour. In how many hours will they meet?
41. GEOMETRY Christie has a cylindrical block that
she needs to paint for an art project.
SOLUTION: Let the distance traveled by one person be x.
So, the distance traveled by the other person be 17.5
– x.
The equation that represents this situation is
.
What is the surface area of the cylinder in square
inches rounded to the nearest square inch? F 1960
G 2413
H 5127
J 6634
Find the time taken by substituting 10 for distance
and 4 for speed.
SOLUTION: The radius of the cylinder is 12 in.
Substitute 20 for h and 12 for r in the formula to find
the volume V of the cylinder.
So, the two people will meet after 2.5 hrs.
ANSWER: 2.5
G is the correct option.
ANSWER: 41. GEOMETRY Christie has a cylindrical block that
she needs to paint for an art project.
G
42. Which of the following is a linear function?
A
B
cylinder in square
inches rounded to the nearest square inch? eSolutions
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What
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area
of the
Page 13
C
ANSWER: ANSWER: C
10-6GThe Binomial Theorem
42. Which of the following is a linear function?
Find the first five terms of each sequence.
43. a 1 = –2, a n + 1 = a n + 5
A
SOLUTION: B
C
D
SOLUTION: The graph of the function
is a straight line.
The first five terms of the sequence are –2, 3, 8, 13
and 18.
ANSWER: So,
is a linear function.
–2, 3, 8, 13, 18
C is the correct option.
ANSWER: C
44. a 1 = 3, a n + 1 = 4a n – 10
SOLUTION: Find the first five terms of each sequence.
43. a 1 = –2, a n + 1 = a n + 5
SOLUTION: The first five terms of the sequence are 3, 2, –2, –18
and –82.
ANSWER: The first five terms of the sequence are –2, 3, 8, 13
and 18.
ANSWER: –2, 3, 8, 13, 18
3, 2, –2, –18, –82
45. a 1 = 4, a n + 1 = 3a n – 6
SOLUTION: eSolutions Manual - Powered by Cognero
44. a 1 = 3, a n + 1 = 4a n – 10
Page 14
ANSWER: ANSWER: 2, –2,
–18, –82Theorem
10-63,The
Binomial
4, 6, 12, 30, 84
Find the sum of each infinite geometric series,
if it exists.
45. a 1 = 4, a n + 1 = 3a n – 6
SOLUTION: 46. SOLUTION: Find the value of r to determine if the sum exists.
The first five terms of the sequence are 4, 6, 12, 30
and 84.
Since
ANSWER: , the sum exists.
4, 6, 12, 30, 84
Use the formula to find the sum.
Find the sum of each infinite geometric series,
if it exists.
46. SOLUTION: Find the value of r to determine if the sum exists.
ANSWER: –4
Since
, the sum exists.
47. Use the formula to find the sum.
SOLUTION: Find the value of r to determine if the sum exists.
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Since
Page 15
, the sum exists.
ANSWER: ANSWER: –4
10-6 The Binomial Theorem
48. 47. SOLUTION: Find the value of r to determine if the sum exists.
SOLUTION: Find the value of r to determine if the sum exists.
Since
, the series diverges and the sum does
not exist.
Since
, the sum exists.
Use the formula to find the sum.
ANSWER: No sum exists.
49. TRAVEL A trip between two towns takes 4 hours
under ideal conditions. The first 150 miles of the trip
is on an interstate, and the last 130 miles is on a
highway with a speed limit that is 10 miles per hour
less than on the interstate.
a. If x represents the speed limit on the interstate,
write expressions for the time spent at that speed and
for the time spent on the other highway.
b. Write and solve an equation to find the speed
limits on the two highways.
SOLUTION: a.
ANSWER: Distance traveled on the interstate is 150 miles.
So, the time spent on the interstate is
Distance traveled on highways is 130 miles.
Speed on highways is x – 10.
So, the time spent on the highways is
48. SOLUTION: Find the value of r to determine if the sum exists.
.
b.
Total time taken for the trip is 4 hrs.
.
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.
Solve for x.
Page 16
ANSWER: b.
sum
exists. Theorem
10-6No
The
Binomial
; 75 mph, 65 mph
49. TRAVEL A trip between two towns takes 4 hours
under ideal conditions. The first 150 miles of the trip
is on an interstate, and the last 130 miles is on a
highway with a speed limit that is 10 miles per hour
less than on the interstate.
a. If x represents the speed limit on the interstate,
write expressions for the time spent at that speed and
for the time spent on the other highway.
State whether each statement is true or false
when n = 1. Explain.
50. SOLUTION: Substitute n = 1 in the equation and simplify.
b. Write and solve an equation to find the speed
limits on the two highways.
SOLUTION: a.
Distance traveled on the interstate is 150 miles.
So, the time spent on the interstate is
So, given statement is true for n = 1.
.
Distance traveled on highways is 130 miles.
Speed on highways is x – 10.
So, the time spent on the highways is
ANSWER: .
b.
Total time taken for the trip is 4 hrs.
true;
51. 3n + 5 is even.
.
Solve for x.
SOLUTION: Substitute n = 1 in the expression and simplify.
So, the speed on the interstate is 75 mph.
Thus, the speed on the highway is 75 – 10 = 65 mph.
Since 8 is an even number, given statement is true
for n = 1.
ANSWER: ANSWER: true; 3(1) + 5 = 8, which is even
a.
2
52. n – 1 is odd.
b.
; 75 mph, 65 mph
State whether each statement is true or false
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when
n = 1.
Explain.
SOLUTION: Substitute n = 1 in the expression and simplify.
Page 17
ANSWER: + 5 = 8,Theorem
which is even
10-6true;
The3(1)
Binomial
2
52. n – 1 is odd.
SOLUTION: Substitute n = 1 in the expression and simplify.
Since 0 is not an odd number, given statement is false
for n = 1.
ANSWER: 2
false; 1 – 1 = 0, which is not odd
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Page 18
Prove that each statement is true for all natural
numbers.
10-7 Proof by Mathematical Induction
Prove that each statement is true for all natural
numbers.
1. 1 + 3 + 5 + … + (2n – 1) = n
1. 1 + 3 + 5 + … + (2n – 1) = n
SOLUTION: Step 1: When n = 1, the left side of the given
2
equation is 1. The right side is 1 or 1, so the equation
is true for n = 1.
2
SOLUTION: Step 1: When n = 1, the left side of the given
2
equation is 1. The right side is 1 or 1, so the equation
is true for n = 1.
Step 2: Assume that 1 + 3 + 5 + . . . + (2k – 1) = k
for some natural number k.
Step 2: Assume that 1 + 3 + 5 + . . . + (2k – 1) = k
for some natural number k.
2
2
2
Step 3: 1 + 3 + 5 + . . . + (2k – 1) + (2(k + 1) – 1)
2
= k + (2(k + 1) – 1)
Step 3: 1 + 3 + 5 + . . . + (2k – 1) + (2(k + 1) – 1)
2
2
= k + (2k + 2 – 1)
2
= k + 2k + 1
2
= (k + 1)
= k + (2(k + 1) – 1)
2
= k + (2k + 2 – 1)
2
= k + 2k + 1
The last expression is the right side of the equation to
be proved, where n = k + 1.
Thus, the equation is true for n = k + 1.
2
= (k + 1)
The last expression is the right side of the equation to
be proved, where n = k + 1.
Thus, the equation is true for n = k + 1.
2
Therefore, 1 + 3 + 5 + . . . + (2n – 1) = n for all
natural numbers n.
2
Therefore, 1 + 3 + 5 + . . . + (2n – 1) = n for all
natural numbers n.
ANSWER: Step 1: When n = 1, the left side of the given
2
equation is 1. The right side is 1 or 1, so the equation
is true for n = 1.
ANSWER: Step 1: When n = 1, the left side of the given
2
equation is 1. The right side is 1 or 1, so the equation
is true for n = 1.
Step 2: Assume that 1 + 3 + 5 + . . . + (2k – 1) = k
for some natural number k.
Step 2: Assume that 1 + 3 + 5 + . . . + (2k – 1) = k
for some natural number k.
2
Step 3: 1 + 3 + 5 + . . . + (2k – 1) + (2(k + 1) – 1)
2
= k + (2(k + 1) – 1)
2
= k + (2k + 2 – 1)
Step 3: 1 + 3 + 5 + . . . + (2k – 1) + (2(k + 1) – 1)
2
= k + 2k + 1
2
= (k + 1)
2
= k + (2(k + 1) – 1)
2
= k + (2k + 2 – 1)
2
= k + 2k + 1
2
= (k + 1)
The last expression is the right side of the equation to
be proved, where n = k + 1. Thus, the equation is
true for n = k + 1. Therefore, 1 + 3 + 5 + . . . + (2n
2
The last expression is the right side of the equation to
be proved, where n = k + 1. Thus, the equation is
true for n = k + 1. Therefore, 1 + 3 + 5 + . . . + (2n
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eSolutions
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= n for
all natural numbers
2
– 1) = n for all natural numbers n.
Page 1
n.
2. The last expression is the right side of the equation to
be proved, where n = k + 1. Thus, the equation is
true for n = k + 1. Therefore, 1 + 3 + 5 + . . . + (2n
2
for Step 2: Assume that
some natural number k.
1) = nbyfor
all natural numbers
n.
10-7–Proof
Mathematical
Induction
Step 3: 1 + 2 + 3 + . . . + k + (k + 1)
2. SOLUTION: Step 1: When n = 1, the left side of the given
equation is 1. The right side is
or 1, so the
The last expression is the right side of the equation to
be proved, where n = k + 1. Thus, the equation is
true for n = k + 1. Therefore,
equation is true for n = 1.
for Step 2: Assume that
some natural number k.
for all natural numbers n.
3. NUMBER THEORY A number is triangular if it
can be represented visually by a triangular array.
Step 3: 1 + 2 + 3 + . . . + k + (k + 1)
a. The first triangular number is 1. Find the next 5
triangular numbers.
The last expression is the right side of the equation to
be proved, where n = k + 1. Thus, the equation is
true for n = k + 1. Therefore,
for all natural numbers n.
b. Write a formula for the nth triangular number.
c. Prove that the sum of the first n triangular
numbers equals
.
ANSWER: Step 1: When n = 1, the left side of the given
equation is 1. The right side is
SOLUTION: a.
or 1, so the
equation is true for n = 1.
Step 2: Assume that
for some natural number k.
The next 5 triangular numbers are 3, 6, 10, 15, 21.
Step 3: 1 + 2 + 3 + . . . + k + (k + 1)
b.
c. Step 1: When n = 1, the left side of the given
equation is
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Page 2
or 1, so the equation is true for n = 1.
The last expression is the right side of the equation to
or 1. The right side is b.
for some c. Step 1: When n = 1, the left side of the given
10-7 Proof by Mathematical Induction
equation is
or 1. The right side is natural number k.
Step 3:
or 1, so the equation is true for n = 1.
Step 2: Assume that
for some natural number k.
The last expression is the right side of the equation to
be proved, where n = k + 1. Thus, the equation is
true for n = k + 1. Therefore,
Step 3:
for all natural numbers n.
Prove that each statement is true for all natural
numbers.
The last expression is the right side of the equation to
be proved, where n = k + 1. Thus, the equation is
true for n = k + 1. Therefore,
for all natural numbers n.
n
4. 10 – 1 is divisible by 9.
SOLUTION: 1
Step 1: 10 – 1 = 9, which is divisible by 9. The
statement is true for n = 1.
k
Step 2: Assume 10 – 1 is divisible by 9 for some
k
ANSWER: a. 3, 6, 10, 15, 21
natural number k. This means that 10 – 1 = 9r for
some whole number r.
Step 3:
b.
c. Step 1: When n = 1, the left side of the given
or 1. The right side is equation is
or 1, so the equation is true for n = 1.
Since r is a whole number, 10r + 1 is a whole
k +1
number. Thus, 10
– 1 is divisible by 9, so the
Step 2: Assume that
n
for some statement is true for n = k + 1. Therefore, 10 – 1 is
divisible by 9 for all natural numbers n.
natural number k.
Step 3:
Step 1: 10 – 1 = 9, which is divisible by 9. The
statement is true for n = 1.
ANSWER: 1
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k
Page 3
Step 2: Assume 10 – 1 is divisible by 9 for some
k
natural number k. This means that 10 – 1 = 9r for
some whole number r.
for all natural numbers n.
ANSWER: 1
1: 10
– 1 = 9, which is
divisible by 9. The
10-7Step
Proof
by Mathematical
Induction
statement is true for n = 1.
k
Step 2: Assume 10 – 1 is divisible by 9 for some
k
natural number k. This means that 10 – 1 = 9r for
some whole number r.
ANSWER: 1
Step 1: 4 – 1 = 3, which is divisible by 3. The
statement is true for n = 1.
k
Step 2: Assume that 4 – 1 is divisible by 3 for some
k
natural number k. This means that 4 – 1 = 3r for
some whole number r.
Step 3:
Step 3:
Since r is a whole number, 10r + 1 is a whole
k +1
number. Thus, 10
– 1 is divisible by 9, so the
n
statement is true for n = k + 1. Therefore, 10 – 1 is
divisible by 9 for all natural numbers n.
Since r is a whole number, 4r + 1 is a whole number.
k +1
Thus, 4
– 1 is divisible by 3, so the statement is
n
true for n = k + 1. Therefore, 4 – 1 is divisible by 3
for all natural numbers n.
n
5. 4 – 1 is divisible by 3.
Find a counterexample to disprove each
statement.
SOLUTION: n
1
Step 1: 4 – 1 = 3, which is divisible by 3. The
statement is true for n = 1.
k
Step 2: Assume that 4 – 1 is divisible by 3 for some
k
natural number k. This means that 4 – 1 = 3r for
some whole number r.
6. 3 + 1 is divisible by 4.
SOLUTION: Test different values of n.
Step 3:
The value n = 2 is a counterexample for the
statement.
ANSWER: n =2
Since r is a whole number, 4r + 1 is a whole number.
k +1
Thus, 4
– 1 is divisible by 3, so the statement is
n
true for n = k + 1. Therefore, 4 – 1 is divisible by 3
for all natural numbers n.
7. 2n + 3n is divisible by 4.
SOLUTION: Test different values of n.
ANSWER: 1
Step 1: 4 – 1 = 3, which is divisible by 3. The
statement is true for n = 1.
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k
Step 2: Assume that 4 – 1 is divisible by 3 for some
k
natural number k. This means that 4 – 1 = 3r for
The value n = 1 is a counterexample for the
ANSWER: n =2
10-7 Proof by Mathematical Induction
The last expression is the right side of the equation to
be proved, where n = k + 1. Thus, the equation is
true for n = k + 1. Therefore,
7. 2n + 3n is divisible by 4.
for all natural numbers
n.
SOLUTION: Test different values of n.
ANSWER: Step 1: When n = 1, the left side of the given
equation is
The value n = 1 is a counterexample for the
statement.
. The right side is
equation is true for n = 1.
or
, so the Step 2: Assume that
ANSWER: n =1
for some natural number k.
CCSS ARGUMENTS Prove that each
statement is true for all natural numbers.
Step 3:
8. SOLUTION: Step 1: When n = 1, the left side of the given
equation is
. The right side is
or
, so the The last expression is the right side of the equation to
equation is true for n = 1.
be proved, where n = k + 1. Thus, the equation is
true for n = k + 1. Therefore,
Step 2: Assume that
for all natural numbers
for some natural n.
number k.
Step 3:
9. SOLUTION: Step 1: When n = 1, the left side of the given
equation is 2. The right side is
The last expression is the right side of the equation to
the equation is true for n = 1.
be proved, where n = k + 1. Thus, the equation is
true for n = k + 1. Therefore,
Step 2: Assume that
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for some natural for all natural numbers
n.
or 2, so number k.
for some natural or 2, so equation is 2. The right side is
number k.
the equation is true for n = 1.
10-7 Proof by Mathematical Induction
Step 3:
Step 2: Assume that
for some natural number k.
Step 3:
The last expression is the right side of the equation to
be proved, where n = k + 1. Thus, the equation is
true for n = k + 1. Therefore,
for all natural numbers n.
The last expression is the right side of the equation to
be proved, where n = k + 1. Thus, the equation is
true for n = k + 1. Therefore,
SOLUTION: for all natural Step 1: When n = 1, the left side of the given
1
equation is 1. The right side is 2 – 1 or 1, so the
equation is true for n = 1.
numbers n.
ANSWER: k –1
Step 1: When n = 1, the left side of the given
equation is 2. The right side is
10. or 2, so Step 2: Assume that 1 + 2 + 4 + . . . + 2
for some natural number k.
k
=2 – 1
Step 3:
the equation is true for n = 1.
Step 2: Assume that
for some natural number k.
Step 3:
The last expression is the right side of the equation to
be proved, where n = k + 1. Thus, the equation is
n–
true for n = k + 1. Therefore, 1 + 2 + 4 + . . . + 2
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1
n
= 2 – 1 for all natural numbers n.
Page 6
= 2k – k + 4k + 4 – 3
2
= 2k + 3k + 1
= (k + 1)(2k + 1)
The last expression is the right side of the equation to
10-7be
Proof
by Mathematical
Induction
proved,
where n = k + 1.
Thus, the equation is
= (k + 1)[2(k + 1) – 1]
n–
true for n = k + 1. Therefore, 1 + 2 + 4 + . . . + 2
1
The last expression is the right side of the equation to
n
= 2 – 1 for all natural numbers n.
be proved, where n = k + 1. Thus, the equation is
ANSWER: true for n = k + 1. Therefore, 1 + 5 + 9 + . . . + (4n
Step 1: When n = 1, the left side of the given
– 3) = n (2n – 1) for all natural numbers n.
1
equation is 1. The right side is 2 – 1 or 1, so the
equation is true for n = 1.
ANSWER: Step 1: When n = 1, the left side of the given
k –1
Step 2: Assume that 1 + 2 + 4 + . . . + 2
for some natural number k.
k
=2 – 1
equation is 1. The right side is 1[2(1) – 1] or 1, so the
equation is true for n = 1.
Step 3:
Step 2: Assume that 1 + 5 + 9 + . . . + (4k – 3) = k
(2k – 1) for some natural number k.
Step 3: 1 + 5 + 9 + . . . + (4k – 3) + [4 (k + 1) – 3]
= k (2k – 1) + [4(k + 1) – 3]
2
= 2k – k + 4k + 4 – 3
2
The last expression is the right side of the equation to
= 2k + 3k + 1
= (k + 1)(2k + 1)
be proved, where n = k + 1. Thus, the equation is
= (k + 1)[2(k + 1) – 1]
n–
true for n = k + 1. Therefore, 1 + 2 + 4 + . . . + 2
1
The last expression is the right side of the equation to
n
= 2 – 1 for all natural numbers n.
be proved, where n = k + 1. Thus, the equation is
true for n = k + 1. Therefore, 1 + 5 + 9 + . . . + (4n
11. – 3) = n (2n – 1) for all natural numbers n.
SOLUTION: Step 1: When n = 1, the left side of the given
equation is 1. The right side is 1[2(1) – 1] or 1, so the
equation is true for n = 1.
12. SOLUTION: Step 2: Assume that 1 + 5 + 9 + . . . + (4k – 3) = k
(2k – 1) for some natural number k.
Step 1: When n = 1, the left side of the given
equation is 1. The right side is
or 1, so Step 3: 1 + 5 + 9 + . . . + (4k – 3) + [4 (k + 1) – 3]
= k (2k – 1) + [4(k + 1) – 3]
the equation is true for n = 1.
Step 2: Assume that
2
= 2k – k + 4k + 4 – 3
2
for some natural
= 2k + 3k + 1
= (k + 1)(2k + 1)
number k.
= (k + 1)[2(k + 1) – 1]
eSolutions
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last expression
is Cognero
the right
Step 3: 1 + 4 + 7 + . . . + (3k – 2) + [3(k + 1) – 2]
side of the equation to
be proved, where n = k + 1. Thus, the equation is
true for n = k + 1. Therefore, 1 + 5 + 9 + . . . + (4n
Page 7
Step 2: Assume that
for some natural
for some natural
number k.
10-7number
Proof by
k. Mathematical Induction
Step 3: 1 + 4 + 7 + . . . + (3k – 2) + [3(k + 1) – 2]
Step 3: 1 + 4 + 7 + . . . + (3k – 2) + [3(k + 1) – 2]
The last expression is the right side of the equation to
The last expression is the right side of the equation to
be proved, where n = k + 1. Thus, the equation is
be proved, where n = k + 1. Thus, the equation is
true for n = k + 1. Therefore,
true for n = k + 1. Therefore,
for all natural for all natural numbers n.
numbers n.
13. ANSWER: Step 1: When n = 1, the left side of the given
equation is 1. The right side is
or 1, so SOLUTION: Step 1: When n = 1, the left side of the given
2
equation is 4(1) – 1 or 3. The right side is 2(1) + 1
or 3, so the equation is true for n = 1.
Step 2: Assume that
the equation is true for n = 1.
Step 2: Assume that 3 + 7 + 11 + . . . + (4k – 1) =
for some natural
2
2k + k for some natural number k.
number k.
Step 3: 3 + 7 + 11 + . . . + (4k – 1) + [4(k + 1) – 1]
Step 3: 1 + 4 + 7 + . . . + (3k – 2) + [3(k + 1) – 2]
= 2k + k + [4(k + 1) – 1]
2
2
= 2k + k + 4k + 3
2
= 2k + 5k + 3
2
= 2k + 4k+2+k+1
2
= [2(k + 1) ] + (k + 1)
The last expression is the right side of the equation to
be proved, where n = k + 1. Thus, the equation is
true for n = k + 1. Therefore, 3 + 7 + 11 + . . . + (4n
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2
– 1) = 2n + n for all natural numbers n.
ANSWER: Page 8
= [2(k + 1) ] + (k + 1)
for some natural
The last expression is the right side of the equation to
10-7be
Proof
by Mathematical
Induction
proved,
where n = k + 1.
Thus, the equation is
true for n = k + 1. Therefore, 3 + 7 + 11 + . . . + (4n
number k.
Step 3:
2
– 1) = 2n + n for all natural numbers n.
ANSWER: Step 1: When n = 1, the left side of the given
2
equation is 4(1) – 1 or 3. The right side is 2(1) + 1
or 3, so the equation is true for n = 1.
Step 2: Assume that 3 + 7 + 11 + . . . + (4k – 1) =
Step 3: 3 + 7 + 11 + . . . + (4k – 1) + [4(k + 1) – 1]
The last expression is the right side of the equation to
2
2k + k for some natural number k.
2
be proved, where n = k + 1. Thus, the equation is
2
true for n = k + 1. Therefore,
= 2k + k + [4(k + 1) – 1]
= 2k + k + 4k + 3
2
for all natural = 2k + 5k + 3
2
= 2k + 4k+2+k+1
numbers n.
2
= [2(k + 1) ] + (k + 1)
The last expression is the right side of the equation to
ANSWER: be proved, where n = k + 1. Thus, the equation is
Step 1: When n = 1, the left side of the given
true for n = k + 1. Therefore, 3 + 7 + 11 + . . . + (4n
2
The right side is
equation is
– 1) = 2n + n for all natural numbers n.
so the equation is true for n = 1.
14. Step 2: Assume that
for some natural
SOLUTION: Step 1: When n = 1, the left side of the given
equation is
. The right side is
number k.
Step 3:
, so the equation is true for n = 1.
Step 2: Assume that
for some natural
number k.
Step 3:
The last expression is the right side of the equation to
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be proved, where n = k + 1. Thus, the equation isPage 9
true for n = k + 1. Therefore,
The last expression is the right side of the equation to
10-7 Proof by Mathematical Induction
be proved, where n = k + 1. Thus, the equation is
The last expression is the right side of the equation to
2
be proved, where n = k + 1. Thus, the equation is
2
(2n – 1)
true for n = k + 1. Therefore,
2
2
true for n = k + 1. Therefore, 1 + 3 + 5 + . . . +
for all natural numbers
n.
for all natural numbers n.
ANSWER: Step 1: When n = 1, the left side of the given
2
equation is 1 or 1. The right side is
15. or 1, so the equation is true for
SOLUTION: n = 1.
Step 1: When n = 1, the left side of the given
2
2
2
2
2
Step 2: Assume that 1 + 3 + 5 + . . . + (2k – 1)
equation is 1 or 1. The right side is
for some natural number k.
or 1, so the equation is true for
n = 1.
Step 3: 1 + 3 + 5 + . . . + (2k – 1) + [2(k + 1) –
2
2
2
2
2
Step 2: Assume that 1 + 3 + 5 + . . . + (2k – 1)
2
2
2
2
1]
for some natural number k.
2
2
2
2
Step 3: 1 + 3 + 5 + . . . + (2k – 1) + [2(k + 1) –
2
1]
The last expression is the right side of the equation to
be proved, where n = k + 1. Thus, the equation is
2
2
2
true for n = k + 1. Therefore, 1 + 3 + 5 + . . . +
2
(2n – 1)
The last expression is the right side of the equation to
n.
be proved, where n = k + 1. Thus, the equation is
2
2
2
true for n = k + 1. Therefore, 1 + 3 + 5 + . . . +
2
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n.
for all natural numbers
for all natural numbers
16. GEOMETRY According to the Interior Angle Sum
Formula, if a convex polygon has n sides, then the
Page 10
sum of the measures of the interior angles of a
polygon equals 180(n – 2). Prove this formula for
using mathematical induction and geometry.
(2n – 1)
for all natural numbers
polygon has 3 sides or 3 vertices, there are 180
degrees. The statement is true for n = 3.
n.
10-7 Proof by Mathematical Induction
Step 2: Assume that the statement is true for
16. GEOMETRY According to the Interior Angle Sum
Formula, if a convex polygon has n sides, then the
sum of the measures of the interior angles of a
polygon equals 180(n – 2). Prove this formula for
using mathematical induction and geometry.
.
Step 3: Consider a convex polygon with n + 1
vertices. Since
, if we take one vertex x
there is another vertex y such that there is one vertex
between x and y in one direction, and n – 2 in the
other direction. Join x and y by a new edge, dividing
our original polygon into two polygons. The new
polygons’ interior angles summed together make up
the sum of the original polygon’s interior angles. One
of the new polygons is a triangle and the other a
SOLUTION: polygon of n vertices (all but the one isolated
Step 1: When n = 3, 180(3 – 2) = 180. When a
between x and y). The triangle has interior angle sum
polygon has 3 sides or 3 vertices, there are 180
180°, and by the inductive hypothesis the other degrees. The statement is true for n = 3.
polygon has interior angle sum
Summing these we get
Step 2: Assume that the statement is true for
.
.
, and the
theorem is proved.
Step 3: Consider a convex polygon with n + 1
vertices. Since
, if we take one vertex x
there is another vertex y such that there is one vertex
between x and y in one direction, and n – 2 in the
Prove that each statement is true for all natural
numbers.
n
17. 5 + 3 is divisible by 4.
other direction. Join x and y by a new edge, dividing
our original polygon into two polygons. The new
SOLUTION: polygons’ interior angles summed together make up
1
the sum of the original polygon’s interior angles. One
Step 1: 5 + 3 = 8, which is divisible by 4. The
statement is true for n = 1.
of the new polygons is a triangle and the other a
polygon of n vertices (all but the one isolated
180°, and by the inductive hypothesis the other Step 2: Assume 5 + 3 is divisible by 4 for some
k
natural number k. This means that 5 + 3 = 4r for
some natural number r.
polygon has interior angle sum
Step 3:
k
between x and y). The triangle has interior angle sum
Summing these we get
.
, and the
theorem is proved.
ANSWER: Step 1: When n = 3, 180(3 – 2) = 180. When a
polygon has 3 sides or 3 vertices, there are 180
degrees. The statement is true for n = 3.
Since r is a natural number, 5r – 3 is a natural
number. Thus, 5
+ 3 is divisible by 4, so the
n
statement is true for n = k + 1. Therefore, 5 + 3 is
divisible by 4 for all natural numbers n.
Step 2: Assume that the statement is true for
k +1
.
Step 3: Consider a convex polygon with n + 1
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vertices. Since
, if we take one vertex x
there is another vertex y such that there is one vertex
ANSWER: 1
Step 1: 5 + 3 = 8, which is divisible by 4. The
Page 11
Since r is a natural number, 5r – 3 is a natural
k +1
number. Thus, 5
+ 3 is divisible by 4, so the
theorem is proved.
n
each statement
is true for all natural
10-7Prove
Proof that
by Mathematical
Induction
numbers.
statement is true for n = k + 1. Therefore, 5 + 3 is
divisible by 4 for all natural numbers n.
n
17. 5 + 3 is divisible by 4.
n
18. 9 – 1 is divisible by 8.
SOLUTION: 1
Step 1: 5 + 3 = 8, which is divisible by 4. The
statement is true for n = 1.
k
Step 2: Assume 5 + 3 is divisible by 4 for some
k
natural number k. This means that 5 + 3 = 4r for
some natural number r.
SOLUTION: 1
Step 1: 9 – 1 = 8, which is divisible by 8. The
statement is true for n = 1.
k
Step 2: Assume that 9 – 1 is divisible by 8 for some
k
natural number k. This means that 9 – 1 = 8r for
some whole number r.
Step 3:
Step 3:
Since r is a natural number, 5r – 3 is a natural
k +1
number. Thus, 5
+ 3 is divisible by 4, so the
n
statement is true for n = k + 1. Therefore, 5 + 3 is
divisible by 4 for all natural numbers n.
Since r is a whole number, 9r + 1 is a whole number.
k +1
Thus, 9
– 1 is divisible by 8, so the statement is
n
true for n = k + 1. Therefore, 9 – 1 is divisible by 8
for all natural numbers n.
ANSWER: 1
Step 1: 5 + 3 = 8, which is divisible by 4. The
statement is true for n = 1.
k
Step 2: Assume 5 + 3 is divisible by 4 for some
k
natural number k. This means that 5 + 3 = 4r for
some natural number r.
Step 3:
ANSWER: 1
Step 1: 9 – 1 = 8, which is divisible by 8. The
statement is true for n = 1.
k
Step 2: Assume that 9 – 1 is divisible by 8 for some
k
natural number k. This means that 9 – 1 = 8r for
some whole number r.
Step 3:
Since r is a natural number, 5r – 3 is a natural
k +1
number. Thus, 5
+ 3 is divisible by 4, so the
n
statement is true for n = k + 1. Therefore, 5 + 3 is
divisible by 4 for all natural numbers n.
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Since r is a whole number, 9r + 1 is a whole number.
k +1
Thus, 9
– 1 is divisible by 8, so the statement is
n
true for n = k + 1. Therefore, 9 – 1 is divisible by 8
for all natural numbers n.
n
18. 9 – 1 is divisible by 8.
Page 12
n
19. 12 + 10 is divisible by 11.
Since r is a whole number, 9r + 1 is a whole number.
k +1
Thus, 9
– 1 is divisible by 8, so the statement is
n
10-7true
Proof
Induction
forby
n =Mathematical
k + 1. Therefore,
9 – 1 is divisible by 8
for all natural numbers n.
Since r is a natural number, 12r – 10 is a natural
k +1
number. Thus, 12
n
+ 10 is divisible by 11, so the
n
19. 12 + 10 is divisible by 11.
statement is true for n = k + 1. Therefore, 12 + 10
is divisible by 11 for all natural numbers n.
SOLUTION: 1
Step 1: 12 + 10 = 22, which is divisible by 11. The
statement is true for n = 1.
k
Step 2: Assume that 12 + 10 is divisible by 11 for
k
some natural number k. This means that 12 + 10 =
11r for some natural number r.
n
20. 13 + 11 is divisible by 12.
SOLUTION: 1
Step 1: 13 + 11 = 24, which is divisible by 12. The
statement is true for n = 1.
k
Step 2: Assume that 13 + 11 is divisible by 12 for
Step 3:
k
some natural number k. This means that 13 + 11 =
12r for some natural number r.
Step 3:
Since r is a natural number, 12r – 10 is a natural
k +1
number. Thus, 12
+ 10 is divisible by 11, so the
n
statement is true for n = k + 1. Therefore, 12 + 10
is divisible by 11 for all natural numbers n.
Since r is a natural number, 13r – 11 is a natural
k +1
number. Thus, 13
ANSWER: + 11 is divisible by 12, so the
n
1
Step 1: 12 + 10 = 22, which is divisible by 11. The
statement is true for n = 1.
k
Step 2: Assume that 12 + 10 is divisible by 11 for
k
some natural number k. This means that 12 + 10 =
11r for some natural number r.
statement is true for n = k + 1. Therefore, 13 + 11
is divisible by 12 for all natural numbers n.
ANSWER: 1
Step 1: 13 + 11 = 24, which is divisible by 12. The
statement is true for n = 1.
k
Step 2: Assume that 13 + 11 is divisible by 12 for
Step 3:
k
some natural number k. This means that 13 + 11 =
12r for some natural number r.
Step 3:
Since r is a natural number, 12r – 10 is a natural
k +1
number.
Thus,
12 by Cognero
+ 10 is divisible by 11, so the
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Page 13
n
statement is true for n = k + 1. Therefore, 12 + 10
is divisible by 11 for all natural numbers n.
Step 2: Assume that 13 + 11 is divisible by 12 for
k
some natural number k. This means that 13 + 11 =
12r for some natural number r.
10-7 Proof by Mathematical Induction
Step 3:
ANSWER: n =3
2
23. n – n + 15 is prime.
SOLUTION: When n = 1, the value of the expression is 15 which
is not a prime number, so the expression is not true
for n = 1.
Since r is a natural number, 13r – 11 is a natural
k +1
number. Thus, 13
ANSWER: + 11 is divisible by 12, so the
n
statement is true for n = k + 1. Therefore, 13 + 11
is divisible by 12 for all natural numbers n.
n =1
2
24. n + n + 23 is prime.
Find a counterexample to disprove each
statement.
SOLUTION: When n = 1, the value of the expression is 25 which
is not a prime number, so the expression is not true
for n = 1.
21. 1 + 2 + 3 + … + n = n
2
SOLUTION: When n = 2, the left side of the given equation is 2.
2
The right side is 2 or 4, so the equation is not true
for n = 2.
ANSWER: n =1
25. NATURE The terms of the Fibonacci sequence are
found in many places in nature. The number of
spirals of seeds in sunflowers is a Fibonacci number,
as is the number of spirals of scales on a pinecone.
The Fibonacci sequence begins 1, 1, 2, 3, 5, 8, … Each element after the first two is found by adding
the previous two terms. If f n stands for the nth
ANSWER: n =2
3
22. 1 + 8 + 27 + … + n = (2n + 2)
2
Fibonacci number, prove that f 1 + f 2 + … + f n = f n +
SOLUTION: 2
When n = 3, the left side of the given equation is 3
3
2
or 27. The right side is 8 or 64, so the equation is not
true for n = 3.
ANSWER: n =3
SOLUTION: Step 1: When n = 1, the left side of the given
equation is f 1. The right side is f 3 – 1. Since f 1 = 1
and f 3 = 2 the equation becomes 1 = 2 – 1 and is true
for n = 1.
2
23. n – n + 15 is prime.
SOLUTION: When n = 1, the value of the expression is 15 which
is notManual
a prime
number,
so the expression is not true
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by Cognero
for n = 1.
– 1.
Step 2: Assume that f 1 + f 2 + … + f k = f k + 2 – 1 for
some natural number k.
Step 3: f 1 + f 2 + … + f k + f k + 1
= f k + 2 – 1 + f k + 1
Page 14
=f k + 1 +f k + 2 – 1
= f k + 3 – 1, since Fibonacci numbers are produced
Step 2: Assume that f 1 + f 2 + … + f k = f k + 2 – 1 for
some natural number k.
10-7 Proof by Mathematical Induction
Step 1: 7 + 5 = 12, which is divisible by 6. The
statement is true for n = 1.
k
Step 2: Assume that 7 + 5 is divisible by 6 for some
k
Step 3: f 1 + f 2 + … + f k + f k + 1
= f k + 2 – 1 + f k + 1
natural number k. This means that 7 + 5 = 6r for
some whole number r.
=f k + 1 +f k + 2 – 1
= f k + 3 – 1, since Fibonacci numbers are produced
Step 3:
by adding the two previous Fibonacci numbers. The
last expression is the right side of the equation to be
proved, where n = k + 1. Thus, the equation is true
for n = k + 1. Therefore, f 1 + f 2 + … + f n = f n + 2 –
1 for all natural numbers n.
ANSWER: Step 1: When n = 1, the left side of the given
equation is f 1. The right side is f 3 – 1. Since f 1 = 1
and f 3 = 2 the equation becomes 1 = 2 – 1 and is true
for n = 1.
Since r is a natural number, 7r – 5 is a natural
k +1
number. Thus, 7
+ 5 is divisible by 6, so the
n
statement is true for n = k + 1. Therefore, 7 + 5 is
divisible by 6 for all natural numbers n.
Step 2: Assume that f 1 + f 2 + … + f k = f k + 2 – 1 for
some natural number k.
ANSWER: 1
Step 1: 7 + 5 = 12, which is divisible by 6. The
statement is true for n = 1.
Step 3: f 1 + f 2 + … + f k + f k + 1
= f k + 2 – 1 + f k + 1
=f k + 1 +f k + 2 – 1
= f k + 3 – 1, since Fibonacci numbers are produced
natural number k. This means that 7 + 5 = 6r for
some whole number r.
by adding the two previous Fibonacci numbers. The
last expression is the right side of the equation to be
Step 3:
k
Step 2: Assume that 7 + 5 is divisible by 6 for some
k
proved, where n = k + 1. Thus, the equation is true
for n = k + 1. Therefore, f 1 + f 2 + … + f n = f n + 2 –
1 for all natural numbers n.
Prove that each statement is true for all natural
numbers or find a counterexample.
n
26. 7 + 5 is divisible by 6.
Since r is a natural number, 7r – 5 is a natural
k +1
number. Thus, 7
SOLUTION: + 5 is divisible by 6, so the
n
1
Step 1: 7 + 5 = 12, which is divisible by 6. The
statement is true for n = 1.
statement is true for n = k + 1. Therefore, 7 + 5 is
divisible by 6 for all natural numbers n.
k
Step 2: Assume that 7 + 5 is divisible by 6 for some
k
natural number k. This means that 7 + 5 = 6r for
some whole number r.
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Step 3:
n
27. 18 – 1 is divisible by 17.
SOLUTION: 1
Page 15
Step 1: 18 – 1 = 17, which is divisible by 17. The
statement is true for n = 1.
n
10-7
Proof
Mathematical
is divisible
by 17. Induction
27. 18
– 1by
Since r is a natural number, 18r + 1 is a natural
k +1
number. Thus, 18
– 1 is divisible by 17, so the
n
SOLUTION: statement is true for n = k + 1. Therefore, 18 – 1 is
divisible by 17 for all natural numbers n.
1
Step 1: 18 – 1 = 17, which is divisible by 17. The
statement is true for n = 1.
2
k
Step 2: Assume that 18 – 1 is divisible by 17 for
28. n + 21n + 7 is a prime number.
k
some natural number k. This means that 18 – 1 =
17r for some natural number r.
SOLUTION: When n = 6, the value of the expression is 169 which
is not a prime number, so the expression is not true
for n = 6.
Step 3:
ANSWER: n =6
2
29. n + 3n + 3 is a prime number.
SOLUTION: When n = 3, the value of the expression is 21 which
is not a prime number, so the expression is not true
for n = 3.
Since r is a natural number, 18r + 1 is a natural
k +1
number. Thus, 18
– 1 is divisible by 17, so the
n
statement is true for n = k + 1. Therefore, 18 – 1 is
divisible by 17 for all natural numbers n.
ANSWER: ANSWER: n =3
1
Step 1: 18 – 1 = 17, which is divisible by 17. The
statement is true for n = 1.
30. k
Step 2: Assume that 18 – 1 is divisible by 17 for
k
some natural number k. This means that 18 – 1 =
17r for some natural number r.
SOLUTION: Step 1: When n = 1, the left side of the given
Step 3:
equation is
or 500. The right side is or 500, so the equation is true for n = 1.
Step 2: Assume that
for some natural number k.
Since r is a natural number, 18r + 1 is a natural
k +1
number. Thus, 18
– 1 is divisible by 17, so the
n
Therefore, 18 – 1 is
divisible by 17 for all natural numbers n.
statement
true forbynCognero
= k + 1.
eSolutions
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- Powered
Step 3:
Page 16
for some that natural number k.
10-7 Proof by Mathematical Induction
for some natural number k.
Step 3:
Step 3:
The last expression is the right side of the equation to
The last expression is the right side of the equation to
be proved, where n = k + 1. Thus, the equation is
be proved, where n = k + 1. Thus, the equation is
true for n = k + 1. Therefore,
true for n = k + 1. Therefore,
for all for all natural numbers n.
natural numbers n.
ANSWER: Step 1: When n = 1, the left side of the given
equation is
31. or 500. The right side
is
or 500, so the equation is true for n =
1.
Step 1: When n = 1, the left side of the given
equation is
Step 2: Assume
that some natural number k.
Step 3:
SOLUTION: for . The right side is
, so the equation is true for n =
1.
Step 2: Assume that
for some natural number k.
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Step 3:
Page 17
natural number k.
Step 3:
10-7for
Proof
bynatural
Mathematical
some
number k.Induction
Step 3:
The last expression is the right side of the equation to
The last expression is the right side of the equation to
be proved, where n = k + 1. Thus, the equation is
be proved, where n = k + 1. Thus, the equation is
true for n = k + 1. Therefore,
true for n = k + 1. Therefore,
or all natural numbers n.
for all natural numbers n.
ANSWER: Step 1: When n = 1, the left side of the given
32. CCSS PERSEVERANCE Refer to the figures
below.
or . The right side is
equation is
or , so the equation is true for n =
1.
a. There is a total of 5 squares in the second figure.
How many squares are there in the third figure?
Step 2: Assume that
for some b. Write a sequence for the first five figures.
natural number k.
c. How many squares are there in a standard
checkerboard?
Step 3:
d. Write a formula to represent the number of
squares in an
grid.
SOLUTION: a. 14
b. 1, 5, 14, 30, 55
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c. 204
Page 18
all natural
numbers n. Induction
10-7for
Proof
by Mathematical
32. CCSS PERSEVERANCE Refer to the figures
below.
33. CHALLENGE Suggest a formula to represent 2 +
4 + 6 + … + 2n, and prove your hypothesis using
mathematical induction.
SOLUTION: Formula to represent 2 + 4 + 6 + … + 2n:
2 + 4 + 6 + … + 2n = 2(1 + 2 + 3 + … + n)
= n(n + 1)
a. There is a total of 5 squares in the second figure.
How many squares are there in the third figure?
Step 1: When n = 1, the left side of the given
b. Write a sequence for the first five figures.
c. How many squares are there in a standard
checkerboard?
equation is 2(1) or 2. The right side is 1(1 + 1) or 2,
so the equation is true for n = 1.
Step 2: Assume that 2 + 4 + 6 + . . . + 2k = k (k + 1)
for some natural number k.
d. Write a formula to represent the number of
squares in an
grid.
Step 3: 2 + 4 + 6 + . . . + 2k + 2(k + 1)
= k(k + 1) + 2(k + 1)
= (k + 1)(k + 2)
SOLUTION: = (k + 1)[(k + 1) + 1]
a. 14
The last expression is the right side of the equation to
b. 1, 5, 14, 30, 55
be proved, where n = k + 1. Thus, the equation is
c. 204
2
true for n = k + 1. Therefore, 2 + 4 + 6 + . . . + n =
n (n + 1) for all natural numbers n.
d.
ANSWER: n(n + 1)
ANSWER: a. 14
b. 1, 5, 14, 30, 55
Step 1: When n = 1, the left side of the given
equation is 2(1) or 2. The right side is 1(1 + 1) or 2,
so the equation is true for n = 1.
c. 204
Step 2: Assume that 2 + 4 + 6 + . . . + 2k = k (k + 1)
for some natural number k.
d.
Step 3: 2 + 4 + 6 + . . . + 2k + 2(k + 1)
= k(k + 1) + 2(k + 1)
33. CHALLENGE Suggest a formula to represent 2 +
4 + 6 + … + 2n, and prove your hypothesis using
mathematical induction.
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= (k + 1)(k + 2)
= (k + 1)[(k + 1) + 1]
The last expression is the right side of the equation to
SOLUTION: be proved, where n = k + 1. Thus, the equationPage
is 19
Formula to represent 2 + 4 + 6 + … + 2n:
true for n = k + 1. Therefore, 2 + 4 + 6 + . . . + n =
2
= k(k + 1) + 2(k + 1)
false, but it is more difficult to prove that it is true.
= (k + 1)(k + 2)
Statements need to be proven true by mathematical
induction, geometrically, or by another method.
= (k + 1)[(k + 1) + 1]
10-7 Proof by Mathematical Induction
The last expression is the right side of the equation to
be proved, where n = k + 1. Thus, the equation is
2
true for n = k + 1. Therefore, 2 + 4 + 6 + . . . + n =
n (n + 1) for all natural numbers n.
35. If a statement is true for n = k and n = k + 1, then it
is also true for n = 1.
SOLUTION: Sample answer: False; assume k = 2, just because a
REASONING Determine whether the following
statements are true or false . Explain.
34. If you cannot find a counterexample to a statement,
then it is true.
statement is true for n = 2 and n = 3 does not mean
that it is true for n = 1.
ANSWER: Sample answer: False; assume k = 2, just because a
SOLUTION: Sample answer: False; Even if a counterexample
cannot immediately be found, it does not mean that
statement is true for n = 2 and n = 3 does not mean
that it is true for n = 1.
one doesn’t exist. A statement can easily be proven
false, but it is more difficult to prove that it is true.
Statements need to be proven true by mathematical
induction, geometrically, or by another method.
36. CHALLENGE Prove
.
SOLUTION: ANSWER: Step 1: When n = 1, the left side of the given
Sample answer: False; Even if a counterexample
3
cannot immediately be found, it does not mean that
equation is1 or 1. The right side is
one doesn’t exist. A statement can easily be proven
so the equation is true for n = 1.
false, but it is more difficult to prove that it is true.
Statements need to be proven true by mathematical
Step 2: Assume that
or 1,
induction, geometrically, or by another method.
for some natural 35. If a statement is true for n = k and n = k + 1, then it
is also true for n = 1.
number k.
Step 3:
SOLUTION: Sample answer: False; assume k = 2, just because a
statement is true for n = 2 and n = 3 does not mean
that it is true for n = 1.
ANSWER: Sample answer: False; assume k = 2, just because a
statement is true for n = 2 and n = 3 does not mean
that it is true for n = 1.
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Page 20
The last expression is the right side of the equation to
36. CHALLENGE Prove
.
be proved, where n = k + 1. Thus, the equation is
true for n = k + 1. Therefore,
for all natural 10-7 Proof by Mathematical Induction
numbers n.
The last expression is the right side of the equation to
be proved, where n = k + 1. Thus, the equation is
3
37. REASONING Find a counterexample to x + 30 >
2
x + 20x.
true for n = k + 1. Therefore,
for all natural SOLUTION: Substitute x = 3 in the given inequality.
numbers n.
ANSWER: Step 1: When n = 1, the left side of the given
So, the inequality is not true for x = 3.
3
equation is1 or 1. The right side is
or 1,
ANSWER: x =3
so the equation is true for n = 1.
Step 2: Assume that
for some natural number k.
38. OPEN ENDED Write a sequence, the formula that
produces it, and determine the formula for the sum of
the terms of the sequence. Then prove the formula
with mathematical induction.
Step 3:
SOLUTION: Sample answer: 6 + 10 + 14 + . . . . The sequence is
produced by a n = 4n + 2. The sum of the sequence is
represented by 2n(n + 2).
Step 1: When n = 1, the left side of the given
equation is 4(1) + 2 or 6. The right side is 2(1)(1 + 2)
or 6, so the equation is true for n = 1.
Step 2: Assume that 6 + 10 + 14 + . . . 4k + 2 = 2k(k
+ 2) for some natural number k.
Step 3: 6 + 10 + 14 + . . . 4k + 2 + 4(k + 1) + 2
= 2k(k + 2) + 4(k + 1) + 2
2
= 2k + 4k + 4k + 4 + 2
2
be proved, where n = k + 1. Thus, the equation is
= 2k + 8k + 6
= 2(k + 1)(k + 3)
= 2(k + 1)[(k + 1) + 2]
true for n = k + 1. Therefore,
numbers n.
The last expression is the right side of the equation to
be proved, where n = k + 1. Thus, the equation is
true for n = k + 1. Therefore, 6 + 10 + 14 + . . . (4n
+ 2) = 2n(n + 2) for all natural numbers n.
The last expression is the right side of the equation to
for all natural eSolutions Manual - Powered by Cognero
Page 21
3
37. REASONING Find a counterexample to x + 30 >
2
x + 20x.
ANSWER: Sample answer: 6 + 10 + 14 + . . . . The sequence is
The last expression is the right side of the equation to
be proved, where n = k + 1. Thus, the equation is
true for n = k + 1. Therefore, 6 + 10 + 14 + . . . (4n
10-7 Proof by Mathematical Induction
+ 2) = 2n(n + 2) for all natural numbers n.
ANSWER: Sample answer: 6 + 10 + 14 + . . . . The sequence is
produced by a n = 4n + 2. The sum of the sequence is
represented by 2n(n + 2).
The last expression is the right side of the equation to
be proved, where n = k + 1. Thus, the equation is
true for n = k + 1. Therefore, 6 + 10 + 14 + . . . (4n
+ 2) = 2n(n + 2) for all natural numbers n.
39. WRITING IN MATH Explain how the concept of
dominoes can help you understand the power of
mathematical induction.
SOLUTION: Step 1: When n = 1, the left side of the given
equation is 4(1) + 2 or 6. The right side is 2(1)(1 + 2)
or 6, so the equation is true for n = 1.
Sample answer: When dominoes are set up, after the
first domino falls, the rest will fall as well. With
induction, once it is proved that the statement is true
Step 2: Assume that 6 + 10 + 14 + . . . 4k + 2 = 2k(k
+ 2) for some natural number k.
for n = 1 (the first domino), n = k (the second
Step 3: 6 + 10 + 14 + . . . 4k + 2 + 4(k + 1) + 2
= 2k(k + 2) + 4(k + 1) + 2
2
domino), and n = k + 1 (the next domino), it will be
true for any integer value (any domino).
= 2k + 4k + 4k + 4 + 2
2
= 2k + 8k + 6
= 2(k + 1)(k + 3)
= 2(k + 1)[(k + 1) + 2]
ANSWER: induction, once it is proved that the statement is true
The last expression is the right side of the equation to
be proved, where n = k + 1. Thus, the equation is
true for n = k + 1. Therefore, 6 + 10 + 14 + . . . (4n
+ 2) = 2n(n + 2) for all natural numbers n.
for n = 1 (the first domino), n = k (the second
39. WRITING IN MATH Explain how the concept of
dominoes can help you understand the power of
mathematical induction.
Sample answer: When dominoes are set up, after the
first domino falls, the rest will fall as well. With
domino), and n = k + 1 (the next domino), it will be
true for any integer value (any domino).
40. WRITING IN MATH Provide a real-world
example other than dominoes that describes
mathematical induction.
SOLUTION: SOLUTION: Sample answer: When dominoes are set up, after the
Sample answer: Climbing a ladder; each step leads to
first domino falls, the rest will fall as well. With
the next step.
induction, once it is proved that the statement is true
for n = 1 (the first domino), n = k (the second
domino), and n = k + 1 (the next domino), it will be
true for any integer value (any domino).
ANSWER: Sample answer: When dominoes are set up, after the
ANSWER: Sample answer: Climbing a ladder; each step leads to
the next step.
41. Which of the following is a counterexample to the
statement below?
first domino falls, the rest will fall as well. With
induction, once it is proved that the statement is true
n + n – 11 is prime.
for n = 1 (the first domino), n = k (the second
domino), and n = k + 1 (the next domino), it will be
true for any integer value (any domino).
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eSolutions
40. WRITING IN MATH Provide a real-world
example other than dominoes that describes
2
A n = -6
B n =4
C n =5
Page 22
ANSWER: Sample answer: Climbing a ladder; each step leads to
nextby
step.
10-7the
Proof
Mathematical Induction
ANSWER: B
41. Which of the following is a counterexample to the
statement below?
2
n + n – 11 is prime.
A n = -6
B n =4
C n =5
D n =6
SOLUTION: When n = 4, the value of the expression is 9 which is
not a prime number, so the expression is not true for
n = 4.
B is the correct option.
42. PROBABILITY Latisha wants to create a 7character password. She wants to use an
arrangement of the first 3 letters of her first name
(lat), followed by an arrangement of the 4 digits in
1986, the year she was born. How many possible
passwords can she create in this way?
F 72
G 144
H 288
J 576
SOLUTION: ANSWER: B
42. PROBABILITY Latisha wants to create a 7character password. She wants to use an
arrangement of the first 3 letters of her first name
(lat), followed by an arrangement of the 4 digits in
1986, the year she was born. How many possible
passwords can she create in this way?
F 72
G is the correct option.
ANSWER: G
43. GRIDDED RESPONSE A gear that is 8 inches in
diameter turns a smaller gear that is 3 inches in
diameter. If the larger gear makes 36 revolutions,
how many revolutions does the smaller gear make in
that time?
G 144
H 288
SOLUTION: Let x be the number of revolutions that small gear
makes.
J 576
The equation that represents the situation is
SOLUTION: .
ANSWER: G is the correct option.
96
ANSWER: G
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Page 23
44. SHORT RESPONSE Write an equation for the
nth line. Show how it fits the pattern for each given
line in the list.
ANSWER: 10-796
Proof by Mathematical Induction
44. SHORT RESPONSE Write an equation for the
nth line. Show how it fits the pattern for each given
line in the list.
Line 1:
Line 2:
Line 3:
Line 4:
Line 5:
Line 4: 4(4 - 1) = 4(3) or 12 and 16 - 4 = 12;
Line 5: 5(5 - 1) = 5(4) or 20 and 25 - 5 = 20
Find the indicated term of each expansion.
45. fourth term of (x + 2y)
6
SOLUTION: Use the Binomial Theorem to write the expansion in
sigma notation.
For the fourth term k = 3.
SOLUTION: ; Line 1: 1(1 - 1) = 1(0) or 0 and 1 - 1 = 0;
Line 2: 2(2 - 1) = 2(1) or 2 and 4 - 2 = 2;
Line 3: 3(3 - 1) = 3(2) or 6 and 9 - 3 = 6;
Line 4: 4(4 - 1) = 4(3) or 12 and 16 - 4 = 12;
Line 5: 5(5 - 1) = 5(4) or 20 and 25 - 5 = 20
ANSWER: 3 3
160x y
46. fifth term of (a + b)
6
ANSWER: ; Line 1: 1(1 - 1) = 1(0) or 0 and 1 - 1 = 0;
Line 2: 2(2 - 1) = 2(1) or 2 and 4 - 2 = 2;
Line 3: 3(3 - 1) = 3(2) or 6 and 9 - 3 = 6;
Line 4: 4(4 - 1) = 4(3) or 12 and 16 - 4 = 12;
Line 5: 5(5 - 1) = 5(4) or 20 and 25 - 5 = 20
Find the indicated term of each expansion.
45. fourth term of (x + 2y)
6
SOLUTION: Use the Binomial Theorem to write the expansion in
sigma notation.
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SOLUTION: Use the Binomial Theorem to write the expansion in
sigma notation.
For the fifth term k = 4.
ANSWER: 2 4
15a b
47. fourth term of (x – y)
9
Page 24
SOLUTION: Use the Binomial Theorem to write the expansion in
ANSWER: ANSWER: 2 4
6 3
–84x y
b by Mathematical Induction
10-715a
Proof
47. fourth term of (x – y)
9
SOLUTION: Use the Binomial Theorem to write the expansion in
sigma notation.
48. BIOLOGY In a particular forest, scientists are
interested in how the population of wolves will
change over the next two years. One model for
animal population is the Verhulst population model,
p n + 1 = p n + rpn (1 – p n), where n represents the
number of time periods that have passed, p n
represents the percent of the maximum sustainable
population that exists at time n, and r is the growth
factor.
For the fourth term k = 3.
a. To find the population of the wolves after one
year, evaluate p 1 = 0.45 + 1.5(0.45)(1 – 0.45).
b. Explain what each number in the expression in
part a represents.
–84x y
c. The current population of wolves is 165. Find the
new population by multiplying 165 by the value in part
a.
ANSWER: 6 3
SOLUTION: a.
48. BIOLOGY In a particular forest, scientists are
interested in how the population of wolves will
change over the next two years. One model for
animal population is the Verhulst population model,
p n + 1 = p n + rpn (1 – p n), where n represents the
number of time periods that have passed, p n
represents the percent of the maximum sustainable
population that exists at time n, and r is the growth
factor.
a. To find the population of the wolves after one
year, evaluate p 1 = 0.45 + 1.5(0.45)(1 – 0.45).
c.
New population = 165 × 0.82125
≈136 wolves
The new population of wolves is about 136.
b.
0.45 represents the maximum sustainable population
of 45% and 1.5 is the growth factor.
b. Explain what each number in the expression in
part a represents.
ANSWER: c. The current population of wolves is 165. Find the
new population by multiplying 165 by the value in part
a.
a. 0.82125
b. 0.45 represents the maximum sustainable
population of 45% and 1.5 is the growth factor.
SOLUTION: a.
c. about 136 wolves
Find the exact solution(s) of each system of
equations.
eSolutions Manual - Powered by Cognero
b.
0.45 represents the maximum sustainable population
of 45% and 1.5 is the growth factor.
49. Page 25
c. about 136 wolves
c. about 136 wolves
10-7 Proof by Mathematical Induction
Find the exact solution(s) of each system of
equations.
Find the exact solution(s) of each system of
equations.
49. 49. SOLUTION: SOLUTION: Replace y by
Replace y by
in the equation .
in the equation .
Substitute the values of x in the equation
Substitute the values of x in the equation
to
find the corresponding y values.
When x = 12:
to
find the corresponding y values.
When x = 12:
When x = 6:
When x = 6:
Therefore, the exact solutions of the system of
equations are (12, –16) and (6, –8).
Therefore, the exact solutions of the system of
equations are (12, –16) and (6, –8).
ANSWER: (6, –8), (12, –16)
ANSWER: (6, –8), (12, –16)
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Page 26
50. ANSWER: ANSWER: (12,Mathematical
–8), by
–16)
10-7(6,
Proof
Induction
Evaluate each expression.
50. 51. P(8, 2)
SOLUTION: Multiply the second equation by –4.
SOLUTION: Add both the equations.
ANSWER: 56
Substitute y = 0 in any one of the original equation
and solve for x.
The solution is
52. P(9, 1)
SOLUTION: .
ANSWER: ANSWER: 9
53. P(12, 6)
Evaluate each expression.
SOLUTION: 51. P(8, 2)
SOLUTION: ANSWER: eSolutions Manual - Powered by Cognero
665,280
ANSWER: 56
54. C(5, 2)
Page 27
ANSWER: ANSWER: 10-79Proof by Mathematical Induction
53. P(12, 6)
70
56. C(20, 17)
SOLUTION: SOLUTION: ANSWER: ANSWER: 1140
665,280
54. C(5, 2)
57. P(12, 2)
SOLUTION: SOLUTION: ANSWER: ANSWER: 10
132
55. C(8, 4)
58. P(7, 2)
SOLUTION: SOLUTION: ANSWER: 70
ANSWER: 42
56. C(20, 17)
eSolutions
Manual - Powered by Cognero
SOLUTION: 59. C(8, 6)
SOLUTION: Page 28
ANSWER: ANSWER: 10-742
Proof by Mathematical Induction
24
59. C(8, 6)
SOLUTION: 62. SOLUTION: ANSWER: 28
ANSWER: 17,640
60. SOLUTION: ANSWER: 1260
61. SOLUTION: ANSWER: 24
62. eSolutions Manual - Powered by Cognero
SOLUTION: Page 29
Practice Test - Chapter 10
ANSWER: 129
1. Find the next 4 terms of the arithmetic sequence 81,
72, 63, … .
MULTIPLE CHOICE What is the eighth term in
the arithmetic sequence that begins 18, 20.2, 22.4,
24.6, …?
SOLUTION: A 26.8
B 29
The common difference is –9.
To find the next term, add –9 to the previous term.
63 –
54 –
45 –
36 –
C 31.2
D33.4
9 = 54
9 = 45
9 = 36
9 = 27
SOLUTION: Given a 1 = 18 and n = 8.
The next 4 terms are 54, 45, 36 and 27.
The common difference is 20.2 – 18 = 2.2.
ANSWER: 54, 45, 36, 27
2. Find the 25th term of an arithmetic sequence for
which a 1 = 9 and d = 5.
Option D is the correct answer.
SOLUTION: Substitute 9, 25 and 5 for a 1, n and d respectively in
ANSWER: D
.
4. Find the four arithmetic means between –9 and 11.
SOLUTION: Given a 1 = –9 and a 6 = 11.
ANSWER: 129
MULTIPLE CHOICE What is the eighth term in
the arithmetic sequence that begins 18, 20.2, 22.4,
24.6, …?
The four arithmetic means between –9 and 11 are –
5, –1, 3 and 7.
ANSWER: –5, –1, 3, 7
B 29
A 26.8
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Manual - Powered by Cognero
C 31.2
5. Find the sum of the arithmetic series for which aPage
1= 1
11, n = 14, and a n = 22.
D33.4
ANSWER: –5, –1,
3, 7- Chapter 10
Practice
Test
ANSWER: 231
5. Find the sum of the arithmetic series for which a 1 =
11, n = 14, and a n = 22.
6. MULTIPLE CHOICE What is the next term in
the geometric sequence below?
SOLUTION: F
G
ANSWER: 231
H
6. MULTIPLE CHOICE What is the next term in
the geometric sequence below?
J
SOLUTION: Find the common ratio (r).
F
G
H
J
The common ratio is
SOLUTION: Find the common ratio (r).
.
Find the next term.
Option H is the correct answer.
ANSWER: H
7. Find the three geometric means between 6 and 1536.
eSolutions Manual - Powered by Cognero
The common ratio is
.
SOLUTION: Given a 1 = 5 and a 6 = 1536.
Page 2
ANSWER: 24, 96, 384
ANSWER: H Test - Chapter 10
Practice
7. Find the three geometric means between 6 and 1536.
8. Find the sum of the geometric series for which a 1 =
15,
SOLUTION: Given a 1 = 5 and a 6 = 1536.
, and n = 5.
SOLUTION: Find r.
The three geometric means between 6 and 1536 are
24, 96 and 384.
ANSWER: 24, 96, 384
8. Find the sum of the geometric series for which a 1 =
15,
, and n = 5.
SOLUTION: ANSWER: Find the sum of each series, if it exists.
9. SOLUTION: There are 12 – 2 + 1 or 11 terms, so n = 11.
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Page 3
ANSWER: ANSWER: does not exist
Practice Test - Chapter 10
Find the sum of each series, if it exists.
11. 45 + 37 + 29 + … + –11
SOLUTION: Given a 1 = 45 and a n = –11.
9. SOLUTION: There are 12 – 2 + 1 or 11 terms, so n = 11.
Find the value of d.
The common difference d is –8.
Find n.
Find the sum.
Find the sum.
Therefore,
.
ANSWER: 220
10. ANSWER: 136
SOLUTION: Since
does not exist.
, the series is diverges and the sum
ANSWER: does not exist
12. SOLUTION: Find the value of r.
11. 45 + 37 + 29 + … + –11
SOLUTION: Given a 1 = 45 and a n = –11.
Since
FindManual
the value
of d. by Cognero
eSolutions
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, the series is convergent.
Find the sum.
Page 4
ANSWER: ANSWER: 136 Test - Chapter 10
Practice
13. Write
12. as a fraction.
SOLUTION: SOLUTION: Find the value of r.
Find the value of r.
Since
, the series is convergent.
Find the sum.
ANSWER: ANSWER: Find the first five terms of each sequence.
13. Write
as a fraction.
SOLUTION: 14. a 1 = –1, a n + 1 = 3a n + 5
SOLUTION: Find the value of r.
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The first five terms of the sequence are –1, 2, 11, 38
Page 5
and 119.
ANSWER: Practice Test - Chapter 10
Find the first five terms of each sequence.
14. a 1 = –1, a n + 1 = 3a n + 5
ANSWER: –1, 2, 11, 38, 119
15. a 1 = 4, a n + 1 = a n + n
SOLUTION: SOLUTION: The first five terms of the sequence are –1, 2, 11, 38
and 119.
The first five terms of the sequence are 4, 5, 7, 10
and 14.
ANSWER: 4, 5, 7, 10, 14
ANSWER: –1, 2, 11, 38, 119
15. a 1 = 4, a n + 1 = a n + n
SOLUTION: 16. MULTIPLE CHOICE What are the first 3 iterates
of f (x) = –5x + 4 for an initial value of x0 = 3?
A 3, –11, 59
B –11, 59, –291
C –1, –6, –11
D 59, –291, 1459
SOLUTION: The first five terms of the sequence are 4, 5, 7, 10
and 14.
ANSWER: 4, 5, 7, 10, 14
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16. MULTIPLE CHOICE What are the first 3 iterates
of f (x) = –5x + 4 for an initial value of x0 = 3?
Option B is the correct answer.
ANSWER: Page 6
ANSWER: 4, 5, 7,
10, 14
Practice
Test
- Chapter 10
ANSWER: 4
3
2 2
3
16a – 96a b + 216a b – 216ab + 81b
16. MULTIPLE CHOICE What are the first 3 iterates
of f (x) = –5x + 4 for an initial value of x0 = 3?
4
18. What is the coefficient of the fifth term of (m + 3n)
6
?
A 3, –11, 59
SOLUTION: Use the Binomial Theorem to write the expansion in
sigma notation.
B –11, 59, –291
C –1, –6, –11
D 59, –291, 1459
SOLUTION: For the fifth term k = 4.
The coefficient of the fifth term is 1215.
ANSWER: 1215
9
19. Find the fourth term of the expansion of (c + d) .
Option B is the correct answer.
ANSWER: B
SOLUTION: Use the Binomial Theorem to write the expansion in
sigma notation.
4
17. Expand (2a – 3b) .
SOLUTION: For the fourth term k = 3.
6 3
ANSWER: 4
3
The fourth term is 84c d .
2 2
3
16a – 96a b + 216a b – 216ab + 81b
4
ANSWER: 18. What is the coefficient of the fifth term of (m + 3n)
6 3
6
84c d
?
SOLUTION: Use Manual
the Binomial
Theorem
to write the expansion in
eSolutions
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by Cognero
sigma notation.
Prove that each statement is true for all
positive integers.
20. .
Page 7
ANSWER: some positive integer k.
6 3
84c d
Practice
Test - Chapter 10
Step 3: Show that the given equation is true for n = k
+1
Prove that each statement is true for all
positive integers.
20. .
SOLUTION: Step 1: When n = 1, the left side of the given
equation is 1. The right side is 1 also, so the equation
is true for n = 1.
for
Step 2: Assume
some positive integer k.
The last expression is the right side of the equation to
be proved, where n = k + 1. Therefore,
for all positive Step 3: Show that the given equation is true for n = k
+1
integers n.
n
21. 11 – 1 is divisible by 10.
SOLUTION: 1
Step 1: 11 – 1 = 10, which is divisible by 10. The
statement is true for n = 1.
k
Step 2: Assume that 11 – 1 is divisible by 10 for
k
some positive integer k. This means that 11 – 1 =
10r for some whole number r.
The last expression is the right side of the equation to
be proved, where n = k + 1. Therefore,
Step 3:
for all positive integers n.
ANSWER: Step 1: When n = 1, the left side of the given
equation is 1. The right side is 1 also, so the equation
is true for n = 1.
Step 2: Assume
for
some positive integer k.
Step 3: Show that the given equation is true for n = k
+1
Since r is a whole umber, 11r + 1 is a whole number.
k +1
Thus, 11
– 1 is divisible by 10, so the statement
n
is true for n = k + 1.Therefore, 11 – 1 is divisible by
10 for all positive integers n.
ANSWER: eSolutions Manual - Powered by Cognero
1
Page 8
Step 1: 11 – 1 = 10, which is divisible by 10. The
statement is true for n = 1.
k +1
Thus, 11
– 1 is divisible by 10, so the statement
n
is true for n = k + 1.Therefore, 11 – 1 is divisible by
10 forTest
all positive
integers
Practice
- Chapter
10 n.
ANSWER: n =1
ANSWER: 1
Step 1: 11 – 1 = 10, which is divisible by 10. The
statement is true for n = 1.
k
Step 2: Assume that 11 – 1 is divisible by 10 for
k
some positive integer k. This means that 11 – 1 =
10r for some whole number r.
23. SCHOOL There are an equal number of girls and
boys in Mr. Marshall’s science class. He needs to
choose 8 students to represent his class at the
science fair. What is the probability that 5 are boys?
SOLUTION: Let b represents the number of boys and g
represents the number of girls.
Step 3:
Sum of the coefficients of the terms = 256
Since r is a whole umber, 11r + 1 is a whole number.
k +1
Thus, 11
– 1 is divisible by 10, so the statement
The term 56b g represents that 5 boys are chosen.
5 3
n
is true for n = k + 1.Therefore, 11 – 1 is divisible by
10 for all positive integers n.
22. Find a counterexample for the following statement.
n
n
The probability that 5 are boys is about 21.9%.
2 + 4 is divisible by 4.
SOLUTION: Since 6 is not divisible by 4, the statement is false
when n = 1.
ANSWER: about 21.9%
24. PENDULUM Laurie swings a pendulum. The
distance traveled per swing decreases by 15% with
each swing. If the pendulum initially traveled 10
inches, find the total distance traveled when the
pendulum comes to a rest.
ANSWER: n =1
23. SCHOOL There are an equal number of girls and
boys in Mr. Marshall’s science class. He needs to
choose 8 students to represent his class at the
science fair. What is the probability that 5 are boys?
SOLUTION: eSolutions Manual - Powered by Cognero
Let b represents the number of boys and g
represents the number of girls.
SOLUTION: Given a 1 = 10 in. and r = 100% – 15% = 0.85
Page 9
ANSWER: aboutTest
21.9%
Practice
- Chapter 10
24. PENDULUM Laurie swings a pendulum. The
distance traveled per swing decreases by 15% with
each swing. If the pendulum initially traveled 10
inches, find the total distance traveled when the
pendulum comes to a rest.
SOLUTION: Given a 1 = 10 in. and r = 100% – 15% = 0.85
Find the sum.
The distance traveled by the pendulum is about 66.7
inches.
ANSWER: about 66.7 inches
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Page 10
Study Guide and Review - Chapter 10
ANSWER: false, sequence
State whether each sentence is true or false . If
false, replace the underlined term to make a
true sentence.
1. An infinite geometric series that has a sum is called a
convergent series.
5. The sum of the first n terms of a series is called the
partial sum.
SOLUTION: True
SOLUTION: True
ANSWER: true
ANSWER: true
2. Mathematical induction is the process of repeatedly
composing a function with itself.
SOLUTION: False, Iteration.
ANSWER: false, iteration
3. The arithmetic means of a sequence are the terms
between any two non-successive terms of an
arithmetic sequence.
SOLUTION: True
ANSWER: true
4. A term is a list of numbers in a particular order.
SOLUTION: False, sequence
ANSWER: false, sequence
5. The sum of the first n terms of a series is called the
partial sum.
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6. The formula a n = a n – 2 + a n – 1 is a recursive
formula.
SOLUTION: True
ANSWER: true
7. A geometric sequence is a sequence in which every
term is determined by adding a constant value to the
previous term.
SOLUTION: False, arithmetic sequence
ANSWER: false, arithmetic sequence
8. An infinite geometric series that does not have a sum
is called a partial sum.
SOLUTION: False, divergent series
ANSWER: false, divergent series
9. Eleven and 17 are two geometric means between 5
and 23 in the sequence 5, 11, 17, 23.
SOLUTION: False, arithmetic means
Page 1
ANSWER: false,
divergent
series - Chapter 10
Study
Guide
and Review
ANSWER: true
9. Eleven and 17 are two geometric means between 5
and 23 in the sequence 5, 11, 17, 23.
Find the indicated term of each arithmetic
sequence.
11. a 1 = 9, d = 3, n = 14
SOLUTION: False, arithmetic means
SOLUTION: Substitute 9 for a 1, 3 for d, and 14 for n in the
ANSWER: false, arithmetic means
formula for the nth term and simplify.
4
10. Using the Binomial Theorem, (x – 2) can be
4
3
2
expanded to x – 8x + 24x – 32x + 16.
SOLUTION: Replace n with 4 in the Binomial Theorem.
ANSWER: 48
12. a 1 = –3, d = 6, n = 22
SOLUTION: Substitute –3 for a 1, 6 for d, and 22 for n in the
The statement is true.
formula for the nth term and simplify.
ANSWER: true
Find the indicated term of each arithmetic
sequence.
11. a 1 = 9, d = 3, n = 14
SOLUTION: Substitute 9 for a 1, 3 for d, and 14 for n in the
formula for the nth term and simplify.
ANSWER: 123
13. a 1 = 10, d = –4, n = 9
SOLUTION: Substitute 10 for a 1, –4 for d, and 9 for n in the
formula for the nth term and simplify.
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ANSWER: 48
Page 2
ANSWER: 123
Study
Guide and Review - Chapter 10
13. a 1 = 10, d = –4, n = 9
SOLUTION: Substitute 10 for a 1, –4 for d, and 9 for n in the
formula for the nth term and simplify.
ANSWER: –86
Find the arithmetic means in each sequence.
15. –12, __, __, __, 8
SOLUTION: Substitute –12 for a 1, 5 for n, and 8 for a 5 in the
formula for the nth term and find d.
ANSWER: –22
14. a 1 = –1, d = –5, n = 18
SOLUTION: Substitute –1 for a 1, –5 for d, and 18 for n in the
The arithmetic means are (–12 + 5) or −7, (–7 + 5)
or −2 and (–2 + 5) or 3.
ANSWER: –7, –2, 3
formula for the nth term and simplify.
16. 15, __, __, 29
SOLUTION: Substitute 15 for a 1, 4 for n, and 29 for a 4 in the
formula for the nth term and find d.
ANSWER: –86
Find the arithmetic means in each sequence.
15. –12, __, __, __, 8
SOLUTION: Substitute –12 for a 1, 5 for n, and 8 for a 5 in the
formula for the nth term and find d.
The arithmetic means are
.
ANSWER: eSolutions Manual - Powered by Cognero
The arithmetic means are (–12 + 5) or −7, (–7 + 5)
17. 12, __, __, __, __, –8
Page 3
ANSWER: Study Guide and Review - Chapter 10
17. 12, __, __, __, __, –8
SOLUTION: Substitute 12 for a 1, 6 for n, and –8 for a 6 in the
formula for the nth term and find d.
ANSWER: 60, 48, 36
19. BANKING Carson saves $40 every 2 months. If he
saves at this rate for two years, how much will he
have at the end of two years?
SOLUTION: Substitute 40 for a 1, 12 for n and 40 for d in the
formula for the nth term and find a 12.
The arithmetic means are (12 – 4) or 8, (8 – 4) or 4,
(4 – 4) or 0 and (0 – 4) or −4.
ANSWER: ANSWER: 8, 4, 0, –4
18. 72, __, __, __, 24
SOLUTION: Substitute 72 for a 1, 5 for n, and 24 for a 5 in the
formula for the nth term and find d.
$480
Find Sn for each arithmetic series.
20. a 1 = 16, a n = 48, n = 6
SOLUTION: Substitute 16 for a 1, 48 for a n, 6 for n in the Sum
formula.
The arithmetic means are (72 – 12) or 60, (60 – 12)
or 48 and (48 – 12) or 36.
ANSWER: 60, 48, 36
19. BANKING Carson saves $40 every 2 months. If he
saves at this rate for two years, how much will he
have at the end of two years?
SOLUTION: Substitute
for a 1,by12Cognero
for n and 40 for d in the
eSolutions
Manual 40
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formula for the nth term and find a 12.
ANSWER: 192
21. a 1 = 8, a n = 96, n = 20
SOLUTION: Substitute 8 for a 1, 96 for a n, 20 for n in the Sum
formula.
Page 4
ANSWER: 192
Study
Guide and Review - Chapter 10
21. a 1 = 8, a n = 96, n = 20
SOLUTION: Substitute 8 for a 1, 96 for a n, 20 for n in the Sum
formula.
ANSWER: 1040
22. 9 + 14 + 19 + … + 74
SOLUTION: Substitute 9 for a 1, 5 for d, and 74 for a n in the
formula for the nth term and find n.
ANSWER: 1040
Substitute 9 for a 1, 74 for a n, 14 for n in the Sum
formula.
22. 9 + 14 + 19 + … + 74
SOLUTION: Substitute 9 for a 1, 5 for d, and 74 for a n in the
formula for the nth term and find n.
ANSWER: 581
23. 16 + 7 + -2 + … + –65
Substitute 9 for a 1, 74 for a n, 14 for n in the Sum
formula.
SOLUTION: Substitute 16 for a 1, –9 for d, and –65 for a n in the
formula for the nth term and find n.
ANSWER: 581
Substitute 16 for a 1, –65 for a n, 10 for n in the Sum
formula.
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23. 16 + 7 + -2 + … + –65
Page 5
ANSWER: ANSWER: –245
581
Study
Guide and Review - Chapter 10
23. 16 + 7 + -2 + … + –65
SOLUTION: Substitute 16 for a 1, –9 for d, and –65 for a n in the
formula for the nth term and find n.
24. DRAMA Laura has a drama performance in 12
days. She plans to practice her lines each night. On
the first night she rehearses her lines 2 times. The
next night she rehearses her lines 4 times. The third
night she rehearses her lines 6 times. On the eleventh
night, how many times has she rehearsed her lines?
SOLUTION: The arithmetic sequence that represents the situation
is 2, 4, 6,…
Substitute 2 for a 1, 2 for d, and 11 for n in the
formula for the nth term and find a 11.
Substitute 16 for a 1, –65 for a n, 10 for n in the Sum
formula.
Substitute 2 for a 1, 22 for a n, 11 for n in the Sum
formula.
ANSWER: –245
24. DRAMA Laura has a drama performance in 12
days. She plans to practice her lines each night. On
the first night she rehearses her lines 2 times. The
next night she rehearses her lines 4 times. The third
night she rehearses her lines 6 times. On the eleventh
night, how many times has she rehearsed her lines?
ANSWER: 132
SOLUTION: The arithmetic sequence that represents the situation
is 2, 4, 6,…
Substitute 2 for a 1, 2 for d, and 11 for n in the
Find the sum of each arithmetic series.
25. formula for the nth term and find a 11.
SOLUTION: Use the formula
.
eSolutions Manual - Powered by Cognero
Substitute 2 for a 1, 22 for a n, 11 for n in the Sum
formula.
There are 17 terms, a 1 = 3(5) – 2 or 13, and a 21 = 3
(21) – 2
or 61.
Page 6
ANSWER: ANSWER: 132
Study
Guide and Review - Chapter 10
319
Find the sum of each arithmetic series.
27. 25. SOLUTION: SOLUTION: Use the formula
Use the formula
.
There are 17 terms, a 1 = 3(5) – 2 or 13, and a 21 = 3
(21) – 2
or 61.
.
There are 9 terms, a 1 = –2(4) + 5 or –3, and a 9 = –2
(12) + 5
or –19.
ANSWER: –99
ANSWER: 629
Find the indicated term for each geometric
sequence.
28. a 1 = 5, r = 2, n = 7
26. SOLUTION: Use the formula
.
SOLUTION: Substitute 5 for a 1, 2 for r, 7 for n in the formula for
the nth term and simplify.
There are 11 terms, a 1 = 6(0) – 1 or –1, and a 11 = 6
(10) – 1
or 59.
The seventh term is 320.
ANSWER: ANSWER: 320
319
29. a 1 = 11, r = 3, n = 3
27. eSolutions Manual - Powered by Cognero
SOLUTION: Use the formula
.
SOLUTION: Page 7
Substitute 11 for a 1, 3 for r, 3 for n in the formula for
the nth term and simplify.
ANSWER: 320
Study
Guide and Review - Chapter 10
29. a 1 = 11, r = 3, n = 3
ANSWER: 8
31. a 8 for
SOLUTION: Substitute 11 for a 1, 3 for r, 3 for n in the formula for
the nth term and simplify.
SOLUTION: Find the common ratio r.
The third term is 99.
Substitute
for a 1, 3 for r, 8 for n in the formula for
the nth term and simplify.
ANSWER: 99
30. SOLUTION: The eighth term is
Substitute 128 for a 1,
for r, 5 for n in the
formula for the nth term and simplify.
.
ANSWER: Find the geometric means in each sequence.
The fifth term is 8.
ANSWER: 8
32. 6, __, __, 162
SOLUTION: Substitute 6 for a 1, 162 for a 4, 4 for n in the formula
for the nth term and find r.
31. a 8 for
SOLUTION: Find the common ratio r.
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The geometric means are 6(3) or 18 and 18(3) orPage
54.8
ANSWER: ANSWER: Study Guide and Review - Chapter 10
72 and 72(±3) or ±216.
ANSWER: Find the geometric means in each sequence.
32. 6, __, __, 162
SOLUTION: Substitute 6 for a 1, 162 for a 4, 4 for n in the formula
34. –4, __, __, 108
SOLUTION: Substitute –4 for a 1, 108 for a 4, 4 for n in the
formula for the nth term and find r.
for the nth term and find r.
The geometric means are –4(–3) or 12 and 12(–3) or
–36.
The geometric means are 6(3) or 18 and 18(3) or 54.
ANSWER: ANSWER: 12, –36
18, 54
33. 8, __, __, __, 648
SOLUTION: Substitute 8 for a 1, 648 for a 5, 5 for n in the formula
for the nth term and find r.
35. SAVINGS Nolan has a savings account with a
current balance of $1500. What would be Nolan’s
account balance after 4 years if he receives 5%
interest annually?
SOLUTION: Substitute 1500 for a 0, 1.05 for r and 4 for n in the
formula
and simplify.
The geometric means are 8(±3) or ±24, ±24(±3) or
72 and 72(±3) or ±216.
ANSWER: ANSWER: $1823.26
Find Sn for each geometric series.
34. –4, __, __, 108
SOLUTION: Substitute –4 for a 1, 108 for a 4, 4 for n in the
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formula for the nth term and find r.
36. a 1 = 15, r = 2, n = 4
SOLUTION: Substitute 15 for a 1, 2 for r, 4 for n in the Sum
formula.
Page 9
ANSWER: ANSWER: $1823.26
Study
Guide and Review - Chapter 10
Find Sn for each geometric series.
36. a 1 = 15, r = 2, n = 4
SOLUTION: Substitute 15 for a 1, 2 for r, 4 for n in the Sum
12,285
38. 5 – 10 + 20 – … to 7 terms
SOLUTION: Find the common ratio r.
formula.
Substitute 5 for a 1, –2 for r, 7 for n in the Sum
formula.
ANSWER: 225
ANSWER: 37. a 1 = 9, r = 4, n = 6
SOLUTION: Substitute 9 for a 1, 4 for r, 6 for n in the Sum
formula.
215
39. 243 + 81 + 27 + … to 5 terms
SOLUTION: Find the common ratio r.
Substitute 243 for a 1,
formula.
ANSWER: for r, 5 for n in the Sum
12,285
38. 5 – 10 + 20 – … to 7 terms
SOLUTION: FindManual
the common
ratio
r.
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ANSWER: ANSWER: 215
Study
Guide and Review - Chapter 10
363
39. 243 + 81 + 27 + … to 5 terms
Evaluate the sum of each geometric series.
40. SOLUTION: Find the common ratio r.
SOLUTION: Substitute 3 for a 1, –2 for r, 7 for n in the Sum
formula.
Substitute 243 for a 1,
for r, 5 for n in the Sum
formula.
ANSWER: 129
41. ANSWER: 363
SOLUTION: Evaluate the sum of each geometric series.
Substitute –1 for a 1,
for r, 8 for n in the Sum
formula.
40. SOLUTION: Substitute 3 for a 1, –2 for r, 7 for n in the Sum
formula.
ANSWER: eSolutions Manual - Powered by Cognero
Page 11
ANSWER: 129
ANSWER: ANSWER: 129
Study
Guide and Review - Chapter 10
41. SOLUTION: Substitute –1 for a 1,
for r, 8 for n in the Sum
42. ADVERTISING Natalie is handing out fliers to
advertise the next student council meeting. She hands
out fliers to 4 people. Then, each of those 4 people
hand out 4 fliers to 4 other people. Those 4 then hand
out 4 fliers to 4 new people. If Natalie is considered
the first round, how many people will have been
given fliers after 4 rounds?
formula.
SOLUTION: Substitute 1 for a 1, 4 for r, 4 for n in the Sum
formula.
ANSWER: Therefore, 85 people will have been given flier after
4 rounds.
ANSWER: 85
42. ADVERTISING Natalie is handing out fliers to
advertise the next student council meeting. She hands
out fliers to 4 people. Then, each of those 4 people
hand out 4 fliers to 4 other people. Those 4 then hand
out 4 fliers to 4 new people. If Natalie is considered
the first round, how many people will have been
given fliers after 4 rounds?
Find the sum of each infinite series, if it exists.
43. SOLUTION: Substitute 1 for a 1, 4 for r, 4 for n in the Sum
formula.
SOLUTION: Substitute 8 for a 1 and
for r in the sum formula.
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Page 12
ANSWER: ANSWER: 85Guide and Review - Chapter 10
Study
does not exist
Find the sum of each infinite series, if it exists.
45. 43. SOLUTION: SOLUTION: Substitute 3 for a 1 and
Substitute 8 for a 1 and
for r in the sum formula.
for r in the sum formula.
ANSWER: ANSWER: 6
32
46. PHYSICAL SCIENCE Maddy drops a ball off of a
building that is 60 feet high. Each time the ball
bounces, it bounces back to
44. its previous height. If the ball continues to follow this pattern, what will be
the total distance that the ball travels?
SOLUTION: Find the value of r to determine if the sum exists.
SOLUTION: Substitute 60 for a 1 and
for r in the sum formula.
Since
, the series diverges and the sum does
not exist.
ANSWER: does not exist
Therefore, the total distance traveled by the ball is 2
(180) – 60 or 300 ft.
ANSWER: 45. eSolutions Manual - Powered by Cognero
SOLUTION: 300 ft
Page 13
ANSWER: ANSWER: 6 Guide and Review - Chapter 10
Study
300 ft
46. PHYSICAL SCIENCE Maddy drops a ball off of a
building that is 60 feet high. Each time the ball
bounces, it bounces back to
its previous height. If the ball continues to follow this pattern, what will be
the total distance that the ball travels?
Find the first five terms of each sequence.
47. a 1 = –3, a n + 1 = a n + 4
SOLUTION: SOLUTION: Substitute 60 for a 1 and
for r in the sum formula.
The first five terms of the sequence are –3, 1, 5, 9,
and 13.
Therefore, the total distance traveled by the ball is 2
(180) – 60 or 300 ft.
ANSWER: ANSWER: –3, 1, 5, 9, 13
300 ft
48. a 1 = 5, a n + 1 = 2a n – 5
Find the first five terms of each sequence.
SOLUTION: 47. a 1 = –3, a n + 1 = a n + 4
SOLUTION: The first five terms of the sequence are 5, 5, 5, 5 and
5.
The first five terms of the sequence are –3, 1, 5, 9,
and 13.
ANSWER: eSolutions Manual - Powered by Cognero
5, 5, 5, 5, 5
ANSWER: –3, 1, 5, 9, 13
49. a 1 = 1, a n + 1 = a n + 5
Page 14
ANSWER: –3, 1, 5, 9, 13
Study
Guide and Review - Chapter 10
48. a 1 = 5, a n + 1 = 2a n – 5
SOLUTION: ANSWER: 1, 6, 11, 16, 21
50. SAVINGS Sari has a savings account with a $12,000
balance. She has a 5% interest rate that is
compounded monthly. Every month Sari adds $500 to
the account. The recursive formula b n = 1.05b n – 1 +
500 describes the balance in Sari’s savings account
after n months. Find the balance of Sari’s account
after 3 months. Round your answer to the nearest
penny.
SOLUTION: a 0 = 12000
The first five terms of the sequence are 5, 5, 5, 5 and
5.
ANSWER: 5, 5, 5, 5, 5
Sari will have $15,467.75 after 3 months.
49. a 1 = 1, a n + 1 = a n + 5
SOLUTION: ANSWER: $15,467.75
Find the first three iterates of each function for
the given initial value.
51. f (x) = 2x + 1, x 0 = 3
SOLUTION: The first five terms of the sequence are 1, 6, 11, 16,
and 21.
ANSWER: 1, 6, 11, 16, 21
50. SAVINGS Sari has a savings account with a $12,000
rate that is
compounded monthly. Every month Sari adds $500 to
the account. The recursive formula b n = 1.05b n – 1 +
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balance.
She
has a 5%
interest
Page 15
ANSWER: $15,467.75
Study
Guide and Review - Chapter 10
Find the first three iterates of each function for
the given initial value.
51. f (x) = 2x + 1, x 0 = 3
ANSWER: 7, 15, 31
52. f (x) = 5x – 4, x 0 = 1
SOLUTION: SOLUTION: The first three iterates are 1, 1, and 1.
The first three iterates are 7, 15, and 31.
ANSWER: 1, 1, 1
ANSWER: 7, 15, 31
52. f (x) = 5x – 4, x 0 = 1
53. f (x) = 6x – 1, x 0 = 2
SOLUTION: SOLUTION: The first three iterates are 11, 65, and 389.
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The first three iterates are 1, 1, and 1.
ANSWER: Page 16
ANSWER: ANSWER: 1, Guide
1, 1 and Review - Chapter 10
Study
53. f (x) = 6x – 1, x 0 = 2
11, 65, 389
54. f (x) = 3x + 1, x 0 = 4
SOLUTION: SOLUTION: The first three iterates are 11, 65, and 389.
The first three iterates are 13, 40, and 121.
ANSWER: ANSWER: 11, 65, 389
13, 40, 121
54. f (x) = 3x + 1, x 0 = 4
SOLUTION: Expand each binomial.
55. (a + b)
3
SOLUTION: Replace n with 3 in the Binomial Theorem.
ANSWER: 3
2
2
a + 3a b + 3ab + b
3
7
56. (y – 3)
SOLUTION: Replace n with 7 in the Binomial Theorem. The first three iterates are 13, 40, and 121.
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eSolutions
ANSWER: Page 17
ANSWER: ANSWER: 3
2
2
3
Study
Review
a Guide
+ 3a band
+ 3ab
+ b - Chapter 10
5
4
3
2
–32z + 240z – 720z + 1080z – 810z + 243
7
56. (y – 3)
58. (4a – 3b)
4
SOLUTION: Replace n with 7 in the Binomial Theorem. SOLUTION: Replace n with 4 in the Binomial Theorem.
ANSWER: 7
6
5
4
3
2
y – 21y + 189y – 945y + 2835y – 5103y +
5103y – 2187
ANSWER: 4
3
2 2
3
256a – 768a b + 864a b – 432ab + 81b
57. (3 – 2z)
4
5
59. SOLUTION: Replace n with 5 in the Binomial Theorem. SOLUTION: Replace n with 5 in the Binomial Theorem.
ANSWER: 5
4
3
2
–32z + 240z – 720z + 1080z – 810z + 243
ANSWER: 58. (4a – 3b)
4
SOLUTION: Replace
n -with
4 inby
theCognero
Binomial Theorem.
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Find the indicated term of each expression.
60. third term of (a + 2b)
8
Page 18
SOLUTION: Use the Binomial Theorem to write the expansion in
ANSWER: ANSWER: 2 5
193,536x y
Study Guide and Review - Chapter 10
Find the indicated term of each expression.
60. third term of (a + 2b)
62. second term of
8
SOLUTION: Use the Binomial Theorem to write the expansion in
sigma notation.
SOLUTION: Use the Binomial Theorem to write the expansion in
sigma notation.
For the second term k = 1.
For the third term k = 2.
ANSWER: ANSWER: -13,107,200x
6 2
112a b
9
61. sixth term of (3x + 4y)
Prove that each statement is true for all
positive integers.
7
SOLUTION: Use the Binomial Theorem to write the expansion in
sigma notation.
63. SOLUTION: Step 1 When n = 1, the left side of the equation is
equal to 2. The right side of the equation is also equal
to 2. So the equation is true for n = 1.
For the sixth term k = 5.
Step 2 Assume that
for some positive integer k.
Step 3
ANSWER: 2 5
193,536x y
62. second term of
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SOLUTION: Use the Binomial Theorem to write the expansion in
sigma notation.
Page 19
positive integer k.
The last expression is the right side of the equation to
Study Guide and Review - Chapter 10
Step 3
be proved, where n = k + 1. Thus, the equation is
true for n = k + 1. Therefore,
for all positive integers n.
n
64. 7 – 1 is divisible by 6.
SOLUTION: The last expression is the right side of the equation to
be proved, where n = k + 1. Thus, the equation is
true for n = k + 1. Therefore,
for all 1
Step 1: When n = 1, 7 – 1 = 7 – 1 or 6. Since 6
divided by 6 is 1, the statement is true for n = 1.
k
Step 2: Assume that 7 – 1 is divisible by 6 for some
k
positive integer k. This means that 7 – 1 = 6r for
some whole number r.
ANSWER: Step 3:
positive integers n.
Step 1 When n = 1, the left side of the equation is
equal to 2. The right side of the equation is also equal
to 2. So the equation is true for n = 1.
Step 2 Assume that
for some positive integer k
Since r is a whole number, 7r + 1 is a whole number.
Step 3:
Thus, 7
k +1
– 1 is divisible by 6, so the statement is
n
true for n = k + 1. Therefore, 7 – 1 is divisible by 6
for all positive integers n.
ANSWER: 1
Step 1: When n = 1, 7 – 1 = 7 – 1 or 6. Since 6
divided by 6 is 1, the statement is true for n = 1.
k
Step 2: Assume that 7 – 1 is divisible by 6 for some
k
positive integer k. This means that 7 – 1 = 6r for
some whole number r.
The last expression is the right side of the equation to
be proved, where n = k + 1. Thus, the equation is
Step 3:
true for n = k + 1. Therefore,
for all eSolutions Manual - Powered by Cognero
positive integers n.
Page 20
Step 2: Assume that 7 – 1 is divisible by 6 for some
k
positive integer k. This means that 7 – 1 = 6r for
some whole number r.
Study
Guide and Review - Chapter 10
Step 3:
ANSWER: 1
Step 1: When n = 1, 5 – 1 = 5 – 1 or 4. Since 4
divided by 4 is 1, the statement is true for n = 1.
k
Step 2: Assume that 5 – 1 is divisible by 4 for some
k
positive integer k. This means that 5 – 1 = 4r for
some whole number r.
Step 3:
Since r is a whole number, 7r + 1 is a whole number.
k +1
Thus, 7
– 1 is divisible by 6, so the statement is
n
true for n = k + 1. Therefore, 7 – 1 is divisible by 6
for all positive integers n.
Since r is a whole number, 5r + 1 is a whole number.
n
65. 5 – 1 is divisible by 4.
k +1
Thus, 5
SOLUTION: true for n = k + 1. Therefore, 5 – 1 is divisible by 4
for all positive integers n.
– 1 is divisible by 4, so the statement is
n
1
Step 1: When n = 1, 5 – 1 = 5 – 1 or 4. Since 4
divided by 4 is 1, the statement is true for n = 1.
Find a counterexample for each statement.
k
Step 2: Assume that 5 – 1 is divisible by 4 for some
k
positive integer k. This means that 5 – 1 = 4r for
some whole number r.
66. 8n + 3 is divisible by 11.
SOLUTION: Substitute n = 2 in the given expression.
Step 3:
19 is not divisible by 11, so the expression is not true
for n = 2.
ANSWER: n =2
Since r is a whole number, 5r + 1 is a whole number.
k +1
Thus, 5
– 1 is divisible by 4, so the statement is
n
true for n = k + 1. Therefore, 5 – 1 is divisible by 4
for all positive integers n.
n+1
67. 6
– 2 is divisible by 17.
SOLUTION: Substitute n = 2 in the given expression.
ANSWER: 1
Step 1: When n = 1, 5 – 1 = 5 – 1 or 4. Since 4
divided by 4 is 1, the statement is true for n = 1.
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k
Step 2: Assume that 5 – 1 is divisible by 4 for some
k
positive integer k. This means that 5 – 1 = 4r for
Page 21
214 is not divisible by 17, so the expression is not true
for n = 2.
ANSWER: n =2
Study
Guide and Review - Chapter 10
n+1
67. 6
– 2 is divisible by 17.
SOLUTION: Substitute n = 2 in the given expression.
214 is not divisible by 17, so the expression is not true
for n = 2.
ANSWER: n =2
2
68. n + 2n + 4 is prime.
SOLUTION: When n = 2, the value of the expression is 12 which
is not a prime number, so the expression is not true
for n = 2.
ANSWER: n =2
69. n + 19 is prime.
SOLUTION: When n = 1, the value of the expression is 20 which
is not a prime number, so the expression is not true
for n = 1.
ANSWER: n =1
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