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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.
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