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Sequences and Series
Arithmetic and Geometric
What is a Sequence?
A Sequence is a list of numbers that follows a pattern.
If the same value is added or subtracted each time, it is
called an arithmetic sequence.
If the same value is multiplied or divided each time, it is
called a geometric sequence.
Some patterns are much easier to determine than others. Here
are some tips that can help with patterns.
If the terms become progressively smaller, subtraction or
division may be involved.
If the terms become progressively larger, addition or
multiplication may be involved.
Describe the pattern that is formed and find the next three terms.
6, 8, 11, 15, 20,
It may help to spread the numbers out and write what was
done to each number to get the next in the gap. This may
help with finding the pattern.
6
8
11
15
20
+2
+3
+4
+5
In each term, the number that is added to the next term
increases by one.
20
26
+6
If the pattern continued, the next term would be 26.
To find a recursive definition for a sequence, you compare each
term to the previous term.
800 -400 200 -100 50
To find the recursive definition for a sequence, first describe the
sequence in words.
It starts with 800.
The terms alternate between positive and negative values
The absolute values of the terms are getting smaller
Each term seems to be the previous term divided by -2.
The next term in the sequence will be -25.
Now, translate the description into the parts of the recursive
formula.
This will be a geometric sequence, since we are dividing each term
1
by -2 (or multiplying by − )
2
𝑎1 = 800
The initial term is 800
1
𝑎𝑛 = 𝑎𝑛−1 ÷ −2, or 𝑎𝑛 = 𝑎𝑛−1 ∗ (− )
2
All this means is that each term is found by dividing the term
before it by -2.
Write a recursive definition for each sequence.
38, 33, 28, 23
Since we are subtracting 5 each time, this is an arithmetic
sequence.
𝑎𝑛 = 𝑎𝑛−1 − 5, where 𝑎1 = 38
All this means is that each term is found by subtracting 5
from the term before it.
The explicit formula for the nth term of an arithmetic sequence is
𝑎𝑛 = 𝑎 + 𝑛 − 1 𝑑
𝑎 is the starting point and 𝑑 is the common difference.
𝑛 is always greater than or equal to 1.
You can write the sequence as 𝑎, 𝑎 + 𝑑, 𝑎 + 2𝑑, 𝑎 + 3𝑑, …
The explicit formula is very helpful because you can find a specific
term without having the know the term immediately before it.
Find the 15th term of an arithmetic sequence whose first three
terms are 20, 16.5, and 13.
First, find 𝑑.
Since the numbers are going down, we are subtracting the
same value every time.
𝑑 = −3.5
𝑎𝑛 = 𝑎 + 𝑛 − 1 𝑑
𝑎15 = 20 + 14 −3.5 = 20 − 49 = −29
The 15th term is this sequence is -29.
To solve word problems that involve arithmetic sequences, identify
the common difference 𝑑, and the number of terms in the
sequence 𝑛.
As a part time home health care aide, you are paid a weekly salary
plus a fixed fuel fee for every patient you visit. You receive $240 in a
week that you visit one patient. You receive $250 in a week that you
visit two patients. How much will you receive if you visit 12 patients
in a week?
Identify 𝑑 by finding the difference between two consecutive terms.
250 - 240 = 10, so the common difference is 10.
Identify the starting value.
You receive $240 for a week with 1 visit.
Identify 𝑛.
You want to know the earnings in a week in which you visit 12
patients.
As a part time home health care aide, you are paid a weekly salary
plus a fixed fuel fee for every patient you visit. You receive $240 in a
week that you visit one patient. You receive $250 in a week that you
visit two patients. How much will you receive if you visit 12 patients
in a week?
𝑑 = 10, 𝑎1 = 240, 𝑛 = 12
𝑎𝑛 = 𝑎 + 𝑛 − 1 𝑑
𝑎12 = 240 + 11 10
𝑎12 = 240 + 110
𝑎12 = 350
You will earn $350 if you visit 12 patients in 1 week.
A geometric sequence has a constant ratio between consecutive
terms. This number is called the common ratio.
A geometric sequence can be described by a recursive formula,
𝑎𝑛 = 𝑎𝑛−1 ∗ 𝑟, or as an explicit formula, 𝑎𝑛 = 𝑎 ∗ 𝑟 𝑛−1 .
Find the 12th term of the geometric sequence 5, 15, 45, ....
Find 𝑟 by finding the common ratio between consecutive terms.
𝑟=3
Substitute 𝑎 = 5 and 𝑟 = 3 into the explicit formula to find a
formula for the nth term of the sequence.
𝑎𝑛 = 𝑎 ∗ 𝑟 𝑛−1
𝑎𝑛 = 5 ∗ 3𝑛−1
𝑎12 = 5 ∗ 311
𝑎12 = 885,735
Find the indicated term of the geometric sequence.
4, 2, 1, ....Find 𝑎10
Since the numbers are going down, we are dividing by 2.
1
This also means that we are multiplying by .
𝑎10 = 4 ∗
1 9
( )
2
Either way, 𝑎10 =
or
1
128
2
4
𝑎𝑛 = 9
2
From 2000 to 2009, your friend's landlord has been allowed to raise
her rent by the same percent each year. In 2000, her rent was
$1000, and in 2003, her rent was $1092.73. What was her rent in
2009?
Identify key information in the problem.
You know that your friend's rent was $1000 in 2000.
This means 𝑎 = 1000.
You also know that her rent in 2003 was $1092.73.
This means that 𝑎4 = 1092.73.
Her rent is raised by the same percent each year, which is
the same as multiplying by a constant.
(e.g. a 5% increase is the same as multiplying by 1.05).
Identify missing information
You need to find the common ratio 𝑟 in order to find the
rent in 2009, 𝑎10 .
From 2000 to 2009, your friend's landlord has been allowed to raise
her rent by the same percent each year. In 2000, her rent was
$1000, and in 2003, her rent was $1092.73. What was her rent in
2009?
Use the explicit formula to find 𝑟.
𝑎𝑛 = 𝑎 ∗ 𝑟 𝑛−1
1092.73 = 1000 ∗ 𝑟 4−1
1.09273 = 𝑟 3
1.03 = 𝑟
Use the value of 𝑟 to find the rent in 2009, 𝑎10 .
𝑎10 = (1000)(1.03)9
𝑎10 ≈ 1304.77
Your friend's rent was $1304.77 in 2009.
What is a Series?
A Series is the sum of a sequence.
Instead of just listing the numbers, we now add them up.
There are two types of series – finite and infinite
A finite series has a set starting point and a set ending point.
For example, the sum of the expression (5 – 2n) from n = 2
to n = 4.
An infinite series has a set starting point, but then goes to
infinity.
For an infinite series, you can only find sum if the absolute
value of the ratio is less than one.
Otherwise, the sum is also infinity.
Summation Notation:
Bottom value is the starting point.
Top value is the ending point.
Expression to be evaluated is to the right.
To do this by hand, we have a formula (on the formula sheet) to
help us with those series that have a lot of terms.
𝑛
𝑆𝑛 = (𝑎1 + 𝑎𝑛 )
2
For this, we need to plug in 2 (the starting point) and 4 (the
ending point) to find 𝑎1 and 𝑎𝑛
5−2 2 =5−4=1
5 − 2 4 = 5 − 8 = −3
3
3
𝑆𝑛 = −3
𝑆𝑛 = (1 − 3)
𝑆𝑛 = (−2)
2
2
Summation Notation:
Fortunately, the TI-84 makes life MUCH easier for us!!
MATH – 0 (zero) gives us the summation notation
Just fill in the boxes and hit enter.
Use x for whatever the variable is.
Summation Notation:
To do this by hand, we need to plug in 1 (the starting point) and 15
(the ending point) to find 𝑎1 and 𝑎𝑛
4 1 −1=4−1=3
4 15 − 1 = 60 − 1 = 59
𝑛 = 15, since there are 15 numbers from 1 to 15.
𝑛
𝑆𝑛 = (𝑎1 + 𝑎𝑛 )
2
15
𝑆𝑛 =
(3 + 59)
2
15
𝑆𝑛 =
(62)
2
𝑆𝑛 = 465
Summation Notation:
On the calculator, we go to Math 0, and enter x = 1 on the bottom,
15 on the top, and 4x – 1 to the right and then press enter.
The series sum is 465.
The debate club is offering a prize at the end of the 10 weeks to a
current member who brings three new members for the first
meeting, and then increases the number of new members they
bring each week by two thereafter. One member qualified for the
prize with the minimum number of new members. How many new
members did the member bring at week 10? For all 10 weeks?
Identify key information in the problem.
A member must bring 3 new members to the first meeting, so
𝑎 = 3.
A member must also bring two more new members to each
meeting, so 𝑑 = 2.
The contest runs for 10 weeks, so 𝑛 = 10.
Identify the information you are trying to find.
You want the 10th term (𝑎10 ), and the sum of the first 10 terms
(𝑠10 ).
The debate club is offering a prize at the end of the 10 weeks to a
current member who brings three new members for the first
meeting, and then increases the number of new members they
bring each week by two thereafter. One member qualified for the
prize with the minimum number of new members. How many new
members did the member bring at week 10? For all 10 weeks?
Use the explicit formula to find 𝑎10 .
𝑎𝑛 = 𝑎 + 𝑛 − 1 𝑑
𝑎10 = 3 + 18 = 21
𝑎10 = 3 + 10 − 1 2 𝑎10 = 3 + 9 2
The member brought 21 new members to the 10th meeting.
Use the value of 𝑎10 to find the total number of new members
brought by the winner.
10
𝑛
(3 + 21) 𝑆10 = 5 24 = 120
𝑆𝑛 = (𝑎1 + 𝑎𝑛 ) 𝑆10 =
2
2
The member brought a total of 120 new members to 10 meetings.
𝑎1 (1−𝑟 𝑛 )
,
1−𝑟
The sum of a finite geometric series is 𝑆𝑛 =
where 𝑎1 is
the first term, 𝑟 is the common ratio, and 𝑛 is the number of terms.
𝑎1
,
1−𝑟
The sum of an infinite geometric series with 𝑟 < 1 is 𝑆 =
where 𝑎1 is the first term and 𝑟 is the common ratio.
If 𝑟 ≥ 1, then the series has no sum (it’s actually infinity).
What is the sum of the first ten terms of the geometric series
8 + 16 + 32 + 64 + 128 + ….?
𝑎1 = 8
𝑟 = 2,
𝑛 = 10
This is a finite series, because we stop at 10 terms.
𝑆𝑛 =
𝑎1 (1−𝑟 𝑛 )
1−𝑟
𝑆10 =
−8184
−1
𝑆10 =
8(1−210 )
1−2
𝑆10 =
8(1−1024)
−1
𝑆10 =
8(−1023)
−1
= 8184 The sum of the first 10 terms is 8184.
Your neighbor hosts a family reunion every year. In 2000, it costs
$1500 to host the reunion. Their expenses have decreased by 10%
per year by asking family members to contribute food and party
supplies.
a) What is the rule for the cost of the family reunion?
b) What was the cost of the reunion in 2005?
c) What was the total cost for hosting the family reunions from
2000 to 2009?
Your neighbor hosts a family reunion every year. In 2000, it costs
$1500 to host the reunion. Their expenses have decreased by 10%
per year by asking family members to contribute food and party
supplies.
a) What is the rule for the cost of the family reunion?
The explicit formula for finding a specific term in a
geometric sequence is 𝑎𝑛 = 𝑎1 𝑟 𝑛−1
𝑎1 = 1500,
𝑟 = 1 − .1 = .9
𝑎𝑛 = (1500)(.9)𝑛−1
Your neighbor hosts a family reunion every year. In 2000, it costs
$1500 to host the reunion. Their expenses have decreased by 10%
per year by asking family members to contribute food and party
supplies.
b) What was the cost of the reunion in 2005?
2005 is the 6th year of the reunion, so 𝑛 = 6.
𝑎𝑛 = (1500)(.9)𝑛−1
𝑎6 = (1500)(.9)5 ≈ 886
Your neighbor hosts a family reunion every year. In 2000, it costs
$1500 to host the reunion. Their expenses have decreased by 10%
per year by asking family members to contribute food and party
supplies.
c) What was the total cost for hosting the family reunions from
2000 to 2009?
To find the total cost of hosting the reunion from 2000 to
2009 (10 years), we can use the formula for a finite
geometric series, 𝑆𝑛 =
𝑆10 =
1500(1−.910 )
;
1−.9
𝑎1 (1−𝑟 𝑛 )
1−𝑟
𝑆10 ≈ 9769.82
The total cost for hosting the reunion from 2000 to 2009
is around $9770.
Your neighbor hosts a family reunion every year. In 2000, it costs
$1500 to host the reunion. Their expenses have decreased by 10%
per year by asking family members to contribute food and party
supplies.
c) What was the total cost for hosting the family reunions from
2000 to 2009?
We can use the Math 0 function on the calculator
instead of using the formula.
We can write this problem in summation notation as
shown here.
Math 0, plug the values in, and get the exact same
answer we got on the previous slide (9769.82)
Either way, the total cost for hosting the reunion from 2000
to 2009 is around $9770.