The table shows the heights of a bungee jumper’s bounces.

The height of the bounces shown in the table above form a geometric sequence. In a geometric sequence, the ratio of successive terms is the same number r, called the common ratio.

I. What is a Geometric Sequence?

Notes 12.2: Geometric Sequences and Series

Ex 1: Find the next three terms in the geometric sequence.

1, 4, 16, 64,…

Step 1 Find the value of r by dividing a term by the one before it. each term by the one before it.

II. Finding Subsequent Terms

Step 2 Multiply each term by 4 to find the next three terms.

64 256 1024 4096

4 4 4

The next three terms are 256, 1024, and 4096.

r=𝑎2

𝑎1=

4

1= 4

Ex 2: Find the next three terms in the geometric sequence.

Step 1

Find the value of r. 𝑟 =𝑎2

𝑎1=

3

−9=

−1

3

Step 2 Multiply each term by to find the next three terms.

The next three terms are

When the terms in a geometric sequence alternate between positive and negative, the value of r is negative.

Helpful Hint

The pattern in the table shows that to get the nth term, multiply the first term by the common ratio raised to the power (n – 1).

The nth term of a geometric sequence with common ratio r and first term a, is

III. Formula for the nth term

Ex 1: The first term of a geometric sequence is 500, and the common ratio is 0.2. What is the 7th term of the sequence?

an = a1rn–1 Write the formula.

a7 = 500(0.2)7–1 Substitute 500 for a1,7 for n, and

0.2 for r.= 500(0.2)6 Simplify the exponent.

= 0.032 Use a calculator.

The 7th term of the sequence is 0.032.

Ex 2: For a geometric sequence, the first term is 5 and the common ratio is 2. Find the 6th term of the sequence?

an = a1rn–1 Write the formula.

a6 = 5(2)6–1 Substitute 5 for a1,6 for n, and 2

for r.= 5(2)5 Simplify the exponent.

= 160 Use a calculator.

The 6th term of the sequence is 160.

Ex 3: If the 9th term of a geometric sequence is 13,122 and the common ratio is -3, find the first term.

an = a1rn–1 Write the formula.

13,122 = a1(–3)9–1 Substitute 13,122 for an, 9 for

n, and –3 for r.13,122= a1 (–3)8

Simplify the exponent.

Use a calculator.

The first term of the sequence is 2.

13,122= a1 (–3)8

13,122= a1 (6561)

2= a1

Ex 4: What is the 8th term of the sequence 1000, 500, 250, 125, …?

an = a1rn–1 Write the formula.

Simplify the exponent.

= 7.8125Use a calculator.

The 8th term of the sequence is 7.8125.

Substitute 1000 for a1,8 for n, and

for r.a8 = 1000( )8–1

r=𝑎2

𝑎1=

500

1000=

1

2

A ball is dropped from a tower. The table shows the heights of the ball’s bounces, which form a geometric sequence. What is the height of the 6th bounce?

Bounce Height (cm)

1 300

2 150

3 75

IV. Application of Geometric Sequence

r=𝑎2

𝑎1=

150

300= .5

an = a1rn–1 Write the formula.

a6 = 300(0.5)6–1 Substitute 300 for a1, 6 for n, and

0.5 for r.

= 300(0.5)5 Simplify the exponent.

= 9.375 Use a calculator.

The height of the 6th bounce is 9.375 cm.

Example 2: A population of fruit flies is growing in such a way that each generation is 1.5 times as large as the last generation. Suppose there were 100 insects in the first generation. How many would there be in the fourth generation?

Solution The populations form a geometric sequence with a1= 100 and r= 1.5 . Use n = 4 in the formula for an..

In the fourth generation, the population is about 338 insects.

an = a1rn–1

a4 = 100(1.5)4–1

a4 = 100(1.5)3

a4 = 337.5

IV. Geometric means

• The terms between any two nonconsecutive terms of a geometric sequence are called geometric means.

• Ex 1. Write a sequence that has two geometric means between 48 and -750. This sequence will have the form 48, ___, ___, -750.

First, find the common ratio.

an = a1rn-1

a4 = 48r4-1

-750 = 48r3

-125/8= r3

-2.5 = r take cube root of both sides

• Now, determine the geometric sequence

• a2= 48(-2.5)

• a2= -120

• a3= -120(-2.5)

• a3= 300

• Sequence:

• 48, -120, 300, -750

• Ex 2. Write a sequence that has two geometric means between 128 and 54.

This sequence will have the form 128, ___, ___, 54.

First, find the common ratio.

an = a1rn-1

a4 = 48r4-1

54= 128r3

27/64= r3

.75 = r take cube root of both sides

• Now, determine the geometric sequence

• a2= 128(.75)

• a2= 96

• a3= 96(.75)

• a3= 72

• Sequence:

• 128, 96, 72, 54

V. Sum of a Finite Geometric Series

• The sum of the first n terms of a finite geometric series is given by

𝑆𝑛 =𝑎1−𝑎1𝑟

𝑛

1−𝑟

• Ex 1: find the sum of the first ten terms of the geometric sequence 16-48+122-432+…

Next, use the formula.

𝑆10 =𝑎1−𝑎1𝑟

𝑛

1−𝑟

𝑆10 =16−16(−3)10

1−(−3)

𝑆10 =16−16(−3)10

1−(−3)

First, find the common ratio𝑟 = 𝑎2 ÷ 𝑎1𝑟 = −48 ÷ 16r=-3

𝑆10 =16−16(59049)

1−(−3)

𝑆10 = −236,192

• Ex 1: find the sum of the first eight terms of the geometric sequence 14-70+350-1750+…

Next, use the formula.

𝑆8 =𝑎1−𝑎1𝑟

𝑛

1−𝑟

𝑆8 =14−14(−5)8

1−(−5)

𝑆8 =14−14(390626)

6

First, find the common ratio𝑟 = 𝑎2 ÷ 𝑎1𝑟 = −70 ÷ 14r=-5

𝑆8 =14−5468750

6

𝑆8 = −911,456

VI. Applications of Geometric Series

• On April 1 of every year for 25 years, Andrea deposits $2000 in an IRA which pays an APR of 10% compounded annually. If she makes no withdrawals, how much will she have in the account at the end of 25 years?

𝑆𝑛 =𝑎1−𝑎1𝑟

𝑛

1−𝑟

𝑆𝑛 =2200 − 2200(1. 10)25

1 − 1.10

𝑆𝑛 =216,363.53

𝑎1 = 2000 ∗ 1.1 = 2200Since we can’t start with zero (we would just get zero as our answer) we start at the end of the first year.

n = 25 We are looking for the end of the 25th year.

r = 1.1 because each year we had what we had before (100% = 1) plus our interest, so 1 + .1 = 1.1