Download Hon Alg 2: Unit 6 SEQUENCES: A sequence is an ordered list of

Hon Alg 2: Unit 6
SEQUENCES: A sequence is an ordered list of numbers.
GENERAL NOTATION: The numbers of a sequence are called TERMS
 “a” refers to the NUMERICAL VALUE of a term in the sequence
 Subscript of “a” refers to the LOCATION of a term in the sequence.
Terms of a sequence generally follow a pattern or operations to move forward to the next term.
a1
a2
a3
a4
a5
a6
…
an
ST
ND
RD
TH
TH
TH
…
NTH
TERM
1
TERM
2
TERM
3
TERM
4
TERM
5
TERM
6
TERM
To move backward between terms, we use OPPOSITE operations
EXAMPLES: For each of the following sequences,
i.
Find a PATTERN to describe the sequence
ii.
Find the NEXT TWO TERMS of the sequence based on your pattern
(1) 5, 8, 11, 14, 17, …
(2) 19, 12, 5, -2, -9, …
(3) 8.3, 10.9, 13.5, 16.1, 18.7, …
(4) 237, 203, 169, 135, 101, …
(5) 23x, 36x, 49x, 62x, 75x, …
(6)
How might you find a19 in each of these sequences?
5
11 7 17 5
, 1,
,
,
,
,…
8
8 4
8 2
ARITHMETIC SEQUENCES: The difference between consecutive terms is constant.
 Patterns of addition or subtraction by a constant value.
 The common difference, d, is constant difference.
o d is POSITIVE when the pattern is _____________________
o d is NEGATIVE when the pattern is _____________________
How do you find the common difference?
For each sequence determine if they are ARITHMETIC and if yes, find the common difference.
(1) -7, 3, 8, 18, 23, …
(3) -23, -31.5, -40, -48.5, …
(2) 19, 27, 35, 43, …
(4) 2, 9, 5, 12, 8, …
How do you find the nth term of an arithmetic sequence?
an = a1 + d(n – 1)
a1 = first term
d = common difference
n – 1 = how many terms a1 and an are apart
(5) a1 = 2, d = 7, n = 14
(6) a1 = 17, d = -9, n = 20
(7) a1 = 35, d = 2.3, n = 8
(8) -17, -13, -9, … a12 = ?
(9) 121, 118, 115, … a21 = ?
(10) 3.46, 3.73, 4, … a17 = ?
Example: Find the missing terms a1 = 3, ______, ______, ______, ______, _______, a7 = 27
i.
What is the total change
in the values of the given
terms?

ii.
How far apart are the
locations of the given
terms?
iii.
What is the common
difference (individual
change between terms)?
Arithmetic Sequences with non-consecutive terms given?
Calculate the value of the common difference, Find the a1 value, Find the an.
(11) a1 = 16 and a6 = 91, a11 = ?
(12) a1 = 10 and a5 = -14, a3 = ?
(13) a3 = - 8 and a8 = 7, a5 = ?
(14) a5 = 9107 and a18 = 11928, a10 = ?
Finding the n for a given value in an arithmetic sequence.
(15) Find n, if an = 633, a1 = 9, and d = 24.
(16) 4, 7, 10, 13, … find n for term = 301
(17) 89, 72, 55, 38, … find n for an = - 115
(18) Find n, if an = 30.06, a1 = 14.38, d = 0.56.
ARITHMETIC SERIES (SUM, Sn)
The sum of consecutive terms for an arithmetic sequence.
EXAMPLE: Sn = sum of the first n terms of a sequence
Find the sum of the arithmetic sequence 4, 11, 18, 25, 32, 39, 46, 53, 60.
FORMULA FOR ARITHMETIC SERIES (SUM):
Sn = a1 + a2 + a3 + a4 +…+ an-1 + an
Sn 
a1  an  n
2
or Alternative Formula: S n 
n
2a1  d ( n  1)
2
PRACTICE:
(19) Find the sum of …
a. the first 100 natural numbers.
b. 23 + 32 + 41 + 50 + 59 + 68 + 77 + 86 + 95 + 104 + 113
c. -3.4 – 1.6 + 0.2 + 2 + 3.8 + 5.6 + 7.4 + 9.2
(20) Find Sn for the described arithmetic series
a. a1 = 25, an = 259, n = 14
b. a1 = 188, an = 34, n = 23 c. a1 = 14.72, an = 24.41, n = 18
(21) Find the sum of the arithmetic sequence
a. S50 of 29, 44, 59, 74, 89, … b. S63 of -19, -13, -7, - 1, … c. S42 of 324, 351, 378, 405, …
(22) Find the sum of the arithmetic series
a. 6 + 27 + 48 + … + 363
b. 17 + 23 + 29 + … + 131
c. 8012 + 7999 + 7986 +…+ 6075
(23) Find n for the given descriptions
a. a1 = 5, an = 77, Sn = 1025
b. a1 = 49, an = -103, Sn = -540
c. a1 = 13.7, an = 164.1, Sn = 1511.3
d. a1 = 5, d = 3, Sn = 440
e. a1 = 4, d = 9, Sn = 58
f. a1 = 234, d = - 18, Sn = 1620
Hon Alg 2: Unit 6
SEQUENCES PART 2
EXAMPLES: For each of the following sequences,
i.
Find a PATTERN to describe the sequence
ii.
Find the NEXT TWO TERMS of the sequence based on your pattern
(1) 3, 6, 12, 24, …
(4) 2x, 6x3, 18x5, 54x7, …
(2) -2, 3, - 4.5, 6.75, …
(5) 6, - 2,
(7) 3.5, 7.35, 15.435, 32.4135, …
2
2
,  ,…
3
9
(3) -4, -20, -100, -500, …
(6)
5
25 125 625
,
,
,
…
4 12
36
108
(8) 7, -8.4, 10.08, -12.096, …
GEOMETRIC SEQUENCES: The consecutive terms differ by a constant factor.
 Patterns of MULTIPICATION by a constant value.
 The COMMON RATIO, r, is constant factor.
How do you find the common ratio?
For each sequence, determine if it is GEOMETRIC and if yes, find the common ratio.
(1) 12, 36, 108, 324, …
(3) 4, -8, 16, - 32, …
(5) 12, 18, 24, 30
(2) 8, 4, 0, -4, …
(4) 20, 30, 45, 67.5, …
(6)
5 5 10 20
, ,
,
2 3 9 27
For geometric cases: How might we find a8 directly? (Hint: Consider manipulating repeated multiplications)
How do you find the nth term of an geometric sequence?
an = a1 (r)n – 1
a1 = first term
n – 1 = how many terms a1 and an are apart
r = common ratio
PRACTICE: Find the indicated nth term and write an equation for the nth term of the sequence.
(7) a1 = 2, r = 5, n = 7
(10) 160, 240, 360, … a9 = ?
(8) a1 =
(11)
1
, r = 3, n = 8
3
125 25 5
,
, , … a6 = ?
32 16 8
(9) a1 = 16807, r = 
3
,n=6
7
(12) 4, -12, 36, … a8 = ?
Example: Find the missing terms a1 = 2.25, ______, ______, ______, a5 = 576
i.
How many times did you do a multiplication
to get between the given terms?
ii.
What is a possible common ratio (individual
change between terms)?
Practice: Find the missing terms between the given terms (Geometric Means)
(13) a1 = -2 and a4 = 54
(14) a1 = 162 and a5 = 32
(15) a1 = 4 and a3 = 100
Geometric Sequences with non-consecutive terms given?
(Step1) Calculate or identify the value of the common ratio (Step 2) Find the a1. (Step 3) Find the an.
(16) a4 = 27, r = 3 , find a7
(17) a3 = 180, r = - 6, find a6
(18) a5  32 2 , r  2 , find a2
(19) a2 = 8, a3 = 2,
find a5
(20) a4 = 192, a6 = 3072,
find a9
(21) a 3  87.5, a5  546.875,
find a1
2
GEOMETRIC SERIES (SUM, Sn): Sum of consecutive terms for a geometric sequence.
(1  r n )
FORMULA: a1 + a2 + a3 + a4 +…+ an-1 + an = S n  a1 (1  r )
PROOF of FORMULA:
Sn = a1 + a2 + a3 + a4 +…+ an-1 + an
every term of the geometric sequence is an = a1 (r)n – 1
Sn = a1 + a1r + a1r2 + a1r3 +…+ a1rn-2 + a1rn-1
Multiply sum by common ratio, r
r • (Sn = a1 + a1(r) + a1(r) 2 + a1(r) 3 +…+ a1(r) n-2 + a1(r) n-1)
Distribute r
2
3
4
n-1
n
rSn = a1(r) + a1(r) + a1(r) + a1(r) + …+ a1(r) + a1(r)
Subtract Sn and rSn
Sn – rSn = a1 – a1(r)n
Combine “cancel” like terms
n
Sn (1 – r) = a1 ( 1 – r )
Factor GCF
(1  r n )
Sn  a1
(1  r )
Solve for Sn
(22) Find the sum of …
a. 3 + 6 + 12 + 24 + 48 + 96 + 192 + 384 =
d. 7 + 21 + 63 + … to 10 terms =
b. 16 + 24 + 36 + 54 + 81 + 121.5 =
e.
1
1
+ + 1 + … to 7 terms =
16
4
c. 12 – 36 + 108 – 324 + 972 – 2916 =
f.
5 10
+ 20 - … to 6 terms =
9
3
(23) Find the indicated geometric series Sn
a. a1 = 3, a8 = 384, r = 2
b. a1 = 162, r = 1/3, n = 6
c. a1 = 1296, an = 1, r = -1/6
e. a1 = 625, r = 3/5, n = 5
f. a2 = -36, a5 = 972, S7 = ?
d. a1 = 5, r = 3, n = 12
(24) Find the first term for each geometric series
a. S8 = 39,360, r = 3, a1 = ?? b. S6 = -364, r = -3, a1 = ??
d. S7 =
381
1
, r = , a1 = ??
64
2
e. S6 =
c. S5 = 249.92, r = 0.2, a1 = ??
3367
3
, r = , a1 = ??
16
4
INFINITE GEOMETRIC SERIES (SUM, Sn): The sum, S, of an infinite
geometric series in which the common ratio is between -1 and 1 ( - 1 ≤ r ≤ 1)
a1
S

S = a1 + a2 + a3 + a4 + … =
(1  r )
PRACTICE: Determine if an infinite geometric series exists or not, if yes find the sum.
 Hint: Find the common ratio
(25) 1 
1 1 1
   ...
2 4 8
(26) 30  20 
40 80
2 2  8  16 2  ...

 ... (27)
3
9
2 1 1 1
...
(28)   
3 3 6 12
2 4 8
(29)   ...
7 7 7
(30) - 3 – 4.5 – 6.75 – …
(31) 243  189  147  ...
(32) -3 – 1.8 – 1.08 - …
(33) – 64 + 48 – 36 + 27 - …