Precalculus 2e — Original English

Geometric Sequences

Learning Objectives

  • Determine if a sequence is geometric (IA 12.3.1).
  • Find the general term (nth term) of a geometric sequence (IA 12.3.2).

Objective 1: Determine if a sequence is geometric (IA 12.3.1)

A sequence is called a geometric sequence if the ratio between consecutive terms is always the same.

The ratio between consecutive terms in a geometric sequence is r, the common ratio, where n is greater than or equal to two.

r=anan-1
This figure shows two sets of sequences where r is the common ratio.
Example 1

Determine if each sequence is geometric. If so, indicate the common ratio.

4,8,16,32,64,128,

−2,6,−12,36,−72,216,

Solution

To determine if the sequence is geometric, we find the ratio of the consecutive terms shown.


Find the ratio ofthe consecutive terms.4,8,16,32,64,128, 841683216643212864 22222 The sequence is geometric. The common ratio isr=2.


Find the ratio ofthe consecutive terms.−2,6,−12,36,−72,216, 6−2−12636−12−7236216−72 −3−2−3−2−3 The sequence is not geometric. There is no common ratio.

Practice Makes Perfect

Determine if each sequence is geometric. If so, indicate the common ratio.

−150, −30, −15, −5, 52, …

.
Find the ratio of consecutive terms.

8, 4, 2, 1, 12, 14, …

.
Find the ratio of consecutive terms.
Example 2

Write the first five terms of the sequence where the first term is 3 and the common ratio is r=−2.

Solution

We start with the first term and multiply it by the common ratio. Then we multiply that result by the common ratio to get the next term, and so on.

a1a2a3a4a5 33·(−2)−6·(−2)12·(−2)−24·(−2) −612−2448

The sequence is 3,−6,12,−24,48,

Practice Makes Perfect

Write the first five terms of the sequence where the first term is 7 and the common ratio is r=-3 .

.
a1 a2 a3 a4 a5
7
The sequence is: ________________________________________

Objective 2: Find the general term (nth term) of a geometric sequence (IA 12.3.2)

Let’s find the formula for the general term of a geometric sequence.

Let’s write the first few terms of the sequence where the first term is a1 and the common ratio is r . We will then look for a pattern.

This figure shows an image of a geometric sequence.
Example 3
Find the general term (nth term) of a geometric sequence.
  • Find the thirteenth term of a sequence where the first term is 81 and the common ratio is r=1/3.
  • Find the ninth term of the sequence 6, 18, 54, 162, 486, 1458, … Then find the general term for the sequence.
Solution
  • Find the thirteenth term of a sequence where the first term is 81 and the common ratio is r=1/3.

    .
    To find the 13th term, use the formula with a1=81, r=1/3 and n=13 an=a1rn-1
    Substitute a13=81(13)13-1
    Simplify a13=81(13)13-1a13=81(13)12a13=1729
  • Find the ninth term of the sequence 6, 18, 54, 162, 486, 1458, … Then find the general term for the sequence.

    .
    Let’s first determine a1 and the common ratio r The first term is 6, so  a1=6The ratio is: 186=5418=16254=482162=1458486=3
    To find the 9th term, use the formula with a1=6, r=3 and n=9.
    Substitute these values and simplify
    an=a1rn-1a9=6(3)9-1a9=6(3)8a9=39366
    To find the general term, substitute a1=6 and r=3 into the formula an=a1rn-1an=6(3)n-1

Practice Makes Perfect

Find the general term (nth term) of a geometric sequence.

Find the sixteenth term of a sequence where the first term is 11 and the common ratio is −6.

Find the 10th term of the sequence 9, 18, 36, 72, 144, 288, …. Then give the formula for the general term.

Many jobs offer an annual cost-of-living increase to keep salaries consistent with inflation. Suppose, for example, a recent college graduate finds a position as a sales manager earning an annual salary of $26,000. He is promised a 2% cost of living increase each year. His annual salary in any given year can be found by multiplying his salary from the previous year by 102%. His salary will be $26,520 after one year; $27,050.40 after two years; $27,591.41 after three years; and so on. When a salary increases by a constant rate each year, the salary grows by a constant factor. In this section, we will review sequences that grow in this way.

Finding Common Ratios

The yearly salary values described form a geometric sequence because they change by a constant factor each year. Each term of a geometric sequence increases or decreases by a constant factor called the common ratio. The sequence below is an example of a geometric sequence because each term increases by a constant factor of 6. Multiplying any term of the sequence by the common ratio 6 generates the subsequent term.

A sequence , {1, 6, 36, 216, 1296, ...} that shows all the numbers have a common ratio of 6.

Example 4

Finding Common Ratios

Is the sequence geometric? If so, find the common ratio.

  1. 1,2,4,8,16,...
  2. 48,12,42,...
Solution

Divide each term by the previous term to determine whether a common ratio exists.

  1. 2 1 =2 4 2 =2 8 4 =2 16 8 =2

    The sequence is geometric because there is a common ratio. The common ratio is 2.

  2. 12 48 = 1 4 4 12 = 1 3 2 4 = 1 2

    The sequence is not geometric because there is not a common ratio.

Analysis

The graph of each sequence is shown in Figure 1. It seems from the graphs that both (a) and (b) appear have the form of the graph of an exponential function in this viewing window. However, we know that (a) is geometric and so this interpretation holds, but (b) is not.

Graph of two sequences where graph (a) is geometric and graph (b) is exponential.
Figure 1

Writing Terms of Geometric Sequences

Now that we can identify a geometric sequence, we will learn how to find the terms of a geometric sequence if we are given the first term and the common ratio. The terms of a geometric sequence can be found by beginning with the first term and multiplying by the common ratio repeatedly. For instance, if the first term of a geometric sequence is a 1 =2 and the common ratio is r=4, we can find subsequent terms by multiplying 24 to get 8 then multiplying the result 84 to get 32 and so on.

a 1 =2 a 2 =(24)=8 a 3 =(84)=32 a 4 =(324)=128

The first four terms are {–2–8–32–128}.

Example 5

Writing the Terms of a Geometric Sequence

List the first four terms of the geometric sequence with a 1 =5 and r=–2.

Solution

Multiply a 1 by 2 to find a 2 . Repeat the process, using a 2 to find a 3 , and so on.

a 1 =5 a 2 =2 a 1 =10 a 3 =2 a 2 =20 a 4 =2 a 3 =40

The first four terms are { 5,–10,20,–40 }.

Using Recursive Formulas for Geometric Sequences

A recursive formula allows us to find any term of a geometric sequence by using the previous term. Each term is the product of the common ratio and the previous term. For example, suppose the common ratio is 9. Then each term is nine times the previous term. As with any recursive formula, the initial term must be given.

Example 6

Using Recursive Formulas for Geometric Sequences

Write a recursive formula for the following geometric sequence.

{6913.520.25...}
Solution

The first term is given as 6. The common ratio can be found by dividing the second term by the first term.

r= 9 6 =1.5

Substitute the common ratio into the recursive formula for geometric sequences and define a 1 .

a n =r a n1 a n =1.5 a n1 for n2 a 1 =6

Analysis

The sequence of data points follows an exponential pattern. The common ratio is also the base of an exponential function as shown in Figure 2

Graph of the geometric sequence.
Figure 2

Using Explicit Formulas for Geometric Sequences

Because a geometric sequence is an exponential function whose domain is the set of positive integers, and the common ratio is the base of the function, we can write explicit formulas that allow us to find particular terms.

a n = a 1 r n1

Let’s take a look at the sequence {183672144288...}. This is a geometric sequence with a common ratio of 2 and an exponential function with a base of 2. An explicit formula for this sequence is

a n =18· 2 n1

The graph of the sequence is shown in Figure 3.

Graph of the geometric sequence.
Figure 3
Example 7

Writing Terms of Geometric Sequences Using the Explicit Formula

Given a geometric sequence with a 1 =3 and a 4 =24, find a 2 .

Solution

The sequence can be written in terms of the initial term and the common ratio r.

3,3r,3 r 2 ,3 r 3 ,...

Find the common ratio using the given fourth term.

a n = a 1 r n1 a 4 =3 r 3 Write the fourth term of sequence in terms of  α 1 and r 24=3 r 3 Substitute 24for a 4 8= r 3 Divide r=2 Solve for the common ratio

Find the second term by multiplying the first term by the common ratio.

a 2 =2 a 1 =2(3) =6

Analysis

The common ratio is multiplied by the first term once to find the second term, twice to find the third term, three times to find the fourth term, and so on. The tenth term could be found by multiplying the first term by the common ratio nine times or by multiplying by the common ratio raised to the ninth power.

Example 8

Writing an Explicit Formula for the n th Term of a Geometric Sequence

Write an explicit formula for the nth term of the following geometric sequence.

{21050250...}
Solution

The first term is 2. The common ratio can be found by dividing the second term by the first term.

10 2 =5

The common ratio is 5. Substitute the common ratio and the first term of the sequence into the formula.

a n = a 1 r (n1) a n =2 5 n1

The graph of this sequence in Figure 4 shows an exponential pattern.

Graph of the geometric sequence.
Figure 4

Solving Application Problems with Geometric Sequences

In real-world scenarios involving geometric sequences, we may need to use an initial term of a 0 instead of a 1 . In these problems, we can alter the explicit formula slightly by using the following formula:

a n = a 0 r n
Example 9

Solving Application Problems with Geometric Sequences

In 2013, the number of students in a small school is 284. It is estimated that the student population will increase by 4% each year.

  1. Write a formula for the student population.
  2. Estimate the student population in 2020.
Solution
  1. The situation can be modeled by a geometric sequence with an initial term of 284. The student population will be 104% of the prior year, so the common ratio is 1.04.

    Let P be the student population and n be the number of years after 2013. Using the explicit formula for a geometric sequence we get

    P n  =284 1.04 n
  2. We can find the number of years since 2013 by subtracting.

    20202013=7

    We are looking for the population after 7 years. We can substitute 7 for n to estimate the population in 2020.

    P 7 =284 1.04 7 374

    The student population will be about 374 in 2020.

Key Equations

..
recursive formula for nth term of a geometric sequence a n =r a n1 ,n 2
explicit formula for nth term of a geometric sequence a n = a 1 r n1

Key Concepts

  • A geometric sequence is a sequence in which the ratio between any two consecutive terms is a constant.
  • The constant ratio between two consecutive terms is called the common ratio.
  • The common ratio can be found by dividing any term in the sequence by the previous term. See Example 4.
  • The terms of a geometric sequence can be found by beginning with the first term and multiplying by the common ratio repeatedly. See Example 5 and Example 7.
  • A recursive formula for a geometric sequence with common ratio r is given by a n =r a n1 for n2 .
  • As with any recursive formula, the initial term of the sequence must be given. See Example 6.
  • An explicit formula for a geometric sequence with common ratio r is given by a n = a 1 r n1 . See Example 8.
  • In application problems, we sometimes alter the explicit formula slightly to a n = a 0 r n . See Example 9.

Section Exercises

Verbal

Exercise 1

What is a geometric sequence?

Solution

A sequence in which the ratio between any two consecutive terms is constant.

Exercise 2

How is the common ratio of a geometric sequence found?

Exercise 3

What is the procedure for determining whether a sequence is geometric?

Solution

Divide each term in a sequence by the preceding term. If the resulting quotients are equal, then the sequence is geometric.

Exercise 4

What is the difference between an arithmetic sequence and a geometric sequence?

Exercise 5

Describe how exponential functions and geometric sequences are similar. How are they different?

Solution

Both geometric sequences and exponential functions have a constant ratio. However, their domains are not the same. Exponential functions are defined for all real numbers, and geometric sequences are defined only for positive integers. Another difference is that the base of a geometric sequence (the common ratio) can be negative, but the base of an exponential function must be positive.

Algebraic

For the following exercises, find the common ratio for the geometric sequence.

Exercise 6

1,3,9,27,81,...

Exercise 7

0.125,0.25,0.5,1,2,...

Solution

The common ratio is 2

Exercise 8

2, 1 2 , 1 8 , 1 32 , 1 128 ,...

For the following exercises, determine whether the sequence is geometric. If so, find the common ratio.

Exercise 9

6,12,24,48,96,...

Solution

The sequence is geometric. The common ratio is 2.

Exercise 10

5,5.2,5.4,5.6,5.8,...

Exercise 11

1, 1 2 , 1 4 , 1 8 , 1 16 ,...

Solution

The sequence is geometric. The common ratio is 1 2 .

Exercise 12

6,8,11,15,20,...

Exercise 13

0.8,4,20,100,500,...

Solution

The sequence is geometric. The common ratio is 5.

For the following exercises, write the first five terms of the geometric sequence, given the first term and common ratio.

Exercise 14

a 1 =8, r=0.3

Exercise 15

a 1 =5, r= 1 5

Solution

5,1, 1 5 , 1 25 , 1 125

For the following exercises, write the first five terms of the geometric sequence, given any two terms.

Exercise 16

a 7 =64, a 10 =512

Exercise 17

a 6 =25, a 8 =6.25

Solution

800,400,200,100,50

For the following exercises, find the specified term for the geometric sequence, given the first term and common ratio.

Exercise 18

The first term is 2, and the common ratio is 3. Find the 5th term.

Exercise 19

The first term is 16 and the common ratio is 1 3 . Find the 4th term.

Solution

a 4 = 16 27

For the following exercises, find the specified term for the geometric sequence, given the first four terms.

Exercise 20

a n ={ 1,2,4,8,... }. Find a 12 .

Exercise 21

a n ={ 2, 2 3 , 2 9 , 2 27 ,... }. Find a 7 .

Solution

a 7 = 2 729

For the following exercises, write the first five terms of the geometric sequence.

Exercise 22

a 1 =486, a n = 1 3 a n1

Exercise 23

a 1 =7, a n =0.2 a n1

Solution

7,1.4,0.28,0.056,0.0112

For the following exercises, write a recursive formula for each geometric sequence.

Exercise 24

a n ={ 1,5,25,125,... }

Exercise 25

a n ={ 32,16,8,4,... }

Solution

a = 1 32, a n = 1 2 a n1

Exercise 26

a n ={ 14,56,224,896,... }

Exercise 27

a n ={ 10,3,0.9,0.27,... }

Solution

a 1 =10, a n =0.3 a n1

Exercise 28

a n ={ 0.61,1.83,5.49,16.47,... }

Exercise 29

a n ={ 3 5 , 1 10 , 1 60 , 1 360 ,... }

Solution

a 1 = 3 5 , a n = 1 6 a n1

Exercise 30

a n ={ 2, 4 3 , 8 9 , 16 27 ,... }

Exercise 31

a n ={ 1 512 , 1 128 , 1 32 , 1 8 ,... }

Solution

a 1 = 1 512 , a n =4 a n1

For the following exercises, write the first five terms of the geometric sequence.

Exercise 32

a n =4 5 n1

Exercise 33

a n =12 ( 1 2 ) n1

Solution

12,6,3, 3 2 , 3 4

For the following exercises, write an explicit formula for each geometric sequence.

Exercise 34

a n ={ 2,4,8,16,... }

Exercise 35

a n ={ 1,3,9,27,... }

Solution

a n = 3 n1

Exercise 36

a n ={ 4,12,36,108,... }

Exercise 37

a n ={ 0.8,4,20,100,... }

Solution

a n =0.8 (5) n1

Exercise 38

a n ={1.25,5,20,80,...}

Exercise 39

a n ={ 1, 4 5 , 16 25 , 64 125 ,... }

Solution

a n = ( 4 5 ) n1

Exercise 40

a n ={ 2, 1 3 , 1 18 , 1 108 ,... }

Exercise 41

a n ={ 3,1, 1 3 , 1 9 ,... }

Solution

a n =3 ( 1 3 ) n1

For the following exercises, find the specified term for the geometric sequence given.

Exercise 42

Let a 1 =4, a n =3 a n1 . Find a 8 .

Exercise 43

Let a n = ( 1 3 ) n1 . Find a 12 .

Solution

a 12 = 1 177,147

For the following exercises, find the number of terms in the given finite geometric sequence.

Exercise 44

a n ={ 1,3,9,...,2187 }

Exercise 45

a n ={ 2,1, 1 2 ,..., 1 1024 }

Solution

There are 12 terms in the sequence.

Graphical

For the following exercises, determine whether the graph shown represents a geometric sequence.

Exercise 46
Graph of a scattered plot with labeled points: (1, -3), (2, -1), (3, 1), (4, 3), and (5, 5). The x-axis is labeled n and the y-axis is labeled a_n.
Exercise 47
Graph of a scattered plot with labeled points: (1, -0.5), (2, 0.25), (3, 1.375), (4, 3.0625), and (5, 5.5938). The x-axis is labeled n and the y-axis is labeled a_n.
Solution

The graph does not represent a geometric sequence.

For the following exercises, use the information provided to graph the first five terms of the geometric sequence.

Exercise 48

a 1 =1, r= 1 2

Exercise 49

a 1 =3, a n =2 a n1

Solution
Graph of a scattered plot with labeled points: (1, 3), (2, 6), (3, 12), (4, 24), and (5, 48). The x-axis is labeled n and the y-axis is labeled a_n.
Exercise 50

a n =27 0.3 n1

Extensions

Exercise 51

Use recursive formulas to give two examples of geometric sequences whose 3rd terms are 200.

Solution

Answers will vary. Examples: a 1 =800, a n =0.5a n1 and a 1 =12.5, a n =4a n1

Exercise 52

Use explicit formulas to give two examples of geometric sequences whose 7th terms are 1024.

Exercise 53

Find the 5th term of the geometric sequence {b,4b,16b,...}.

Solution

a 5 =256b

Exercise 54

Find the 7th term of the geometric sequence {64a(b),32a(3b),16a(9b),...}.

Exercise 55

At which term does the sequence {10,12,14.4,17.28,...} exceed 100?

Solution

The sequence exceeds 100 at the 14th term, a 14 107.

Exercise 56

At which term does the sequence { 1 2187 , 1 729 , 1 243 , 1 81 ... } begin to have integer values?

Exercise 57

For which term does the geometric sequence a n =36 ( 2 3 ) n1 first have a non-integer value?

Solution

a 4 = 32 3 is the first non-integer value

Exercise 58

Use the recursive formula to write a geometric sequence whose common ratio is an integer. Show the first four terms, and then find the 10th term.

Exercise 59

Use the explicit formula to write a geometric sequence whose common ratio is a decimal number between 0 and 1. Show the first 4 terms, and then find the 8th term.

Solution

Answers will vary. Example: Explicit formula with a decimal common ratio: a n =400 0.5 n1 ; First 4 terms: 400,200,100,50; a 8 =3.125

Exercise 60

Is it possible for a sequence to be both arithmetic and geometric? If so, give an example.

common ratio
the ratio between any two consecutive terms in a geometric sequence
geometric sequence
a sequence in which the ratio of a term to a previous term is a constant