Example

Solution

For each geometric sequence, find the common ratio.

a) -2, -12, -72, -432, . . .

b) 50, 10, 2, 0.4, 0.08, . . .

Sequence Common Ratio

a) -2, -12, -72, -432, . . .

b) 50, 10, 2, 0.4, 0.08, . . .

12

26,

72

126, and so on.

10

500.2,

2

100.2, and so on.

r = 6

r = 0.2

Example

Solution

Find the 8th term of each sequence.

a) –2, –12, –72, –432, –2592, . . .

b) 50, 10, 2, 0.4, 0.08, . . .

a) First, we note that a1 = –2, n = 8, and r = 6.

The formula

an = a1rn – 1

gives us

a8 = –2·68 – 1 = –2·67 = –2(279936) = –559872.

Solution continued

b) First, we note that a1 = 50, n = 8, and r = 0.2.

The formula

an = a1rn – 1

gives us

a8 = 50·(0.2)8 – 1 = 50·(0.2)7 = 50(0.0000128)

= 0.00064.

Sum of the First n Terms of a Geometric Sequence

Example

Solution

Find the sum of the first 9 terms of the geometric sequence -1, 4, -16, 64, . . . .

First, we note that

a1 = -1, n = 9, and 4

4.1

r

91(1 ( 4) ) 1(1 262,144) 262,14552,429.

9 1 ( 4) 5 5S

1(1 )

1

n

n

a rS

r

Then, substituting in the formulawe have

Example

Solution

Determine whether each series has a limit. If a limit exists, find it.

a) -2 – 12 – 72 – 4323 – · · ·

b) 50 + 10 + 2 + 0.4 + 0.08 + · · ·

a) Here r = 6, so | r | = | 6 | = 6. Since | r | > 1, the series does not have a limit.

Solution continued

We find the limit by substituting into the formula for S∞:

50 5062.5.

1 0.2 0.8S

b) Here r = 0.2, so | r | = | 0.2 | = 0.2. Since | r | < 1, the series does have a limit.

Example

SolutionFind the fraction notation for 0.482482482….

We can express this as

0.482 + 0.000482 + 0.000000482 + · · ·.

This is an infinite geometric series, where a1 = 0.482 and r = 0.001. Since | r | < 1, this series has a limit:

Thus fraction notation for 0.482482482… is 482.

999

0.482 .482 482

1 0.001 .999 999S

Problem Solving

For some problem-solving situations, the translation may involve geometric sequences or series.