# Infinite Geometric Series Worksheet

Infinite Geometric Series Worksheet
• Page 1
1.
Find the sum of the infinite geometric series.
$\frac{\sqrt{5}}{\sqrt{5}+1}+\frac{\sqrt{5}}{5+\sqrt{5}}+\frac{\sqrt{5}}{5\sqrt{5}+5}+....$
 a. 1.25 b. does not exist c. 2.25 d. 3.25

#### Solution:

r = a2a1=55+555+1
[Common ratio.]

= 55+5×5+15

= 55(5+1)×5+15=15
[Simplify.]

|r| = |15| < 1, so the series converges and sum exists.

S = a11-r
[Sum of infinite terms of geometric series.]

= 55+11-15

= 55+1×55-1 = 55-1 = 1.25
[Simplify.]

So, the sum of the infinite geometric series is 1.25.

Correct answer : (1)
2.
What is the sum of the infinite geometric series 5 + 2 + 0.8 + ...?
 a. $\frac{25}{7}$ b. 7.8 c. does not exist d. $\frac{25}{3}$

#### Solution:

r = a2a1 = 2 / 5 = 0.4
[Replace a1 = 5 and a2 = 2.]

| r | = | 0.4 | = 0.4 < 1, so the series converges and the sum exists.

S = a11 - r = 5 / 1-0.4
[Use the formula and replace a1 = 5, r = 0.4.]

S = 5 / 0.6 = 25 / 3
[Simplify.]

So, the sum of the infinite geometric series is 25 / 3.

Correct answer : (4)
3.
The sum of an infinite geometric series is 252 and the common ratio is $\frac{1}{7}$. Find the first term.
 a. 288 b. 216 c. 7 d. 36

#### Solution:

S = a11-r
[Sum of an infinite geometric series.]

252 = a11-17
[Replace S = 252, r = 1 / 7.]

252(1 - 1 / 7) = a1
[Cross multiply.]

252 - 36 = a1
[Simplify.]

216 = a1

So, the first term of an infinite geometric series is 216.

Correct answer : (2)
4.
Find the sum of the infinite geometric series.
48 + 480 + 4800 + ......
 a. does not exist b. $\frac{16}{3}$ c. 5328 d. 48000

#### Solution:

r = a2a1 = 480 / 48 = 10
[Replace a1 = 48 and a2 = 480.]

Since | r | = | 10 | = 10 > 1, the series diverges, so the sum does not exist.

Correct answer : (1)
5.
Find the sum of the infinite geometric series.
- 196 - 28 - 4 - ......
 a. - $\frac{686}{3}$ b. Does not exist c. - 228 d. - 168

#### Solution:

r = a2a1 = - 28- 196 = 1 / 7
[Replace a2 = - 28, a1 = - 196.]

| r | = | 1 / 7 | = 1 / 7 < 1, so the series converges and the sum exists.

S = a11Ã¢â‚¬â€œ r = - 1961-17
[Use the formula, and replace a1 = - 196, r = 1 / 7.]

S = - 196(7 / 6)
[Simplify.]

S = - 686 / 3

So, the sum of the infinite geometric series is - 686 / 3.

Correct answer : (1)
6.
Simplify:
 a. $\frac{12}{143}$ b. $\frac{12}{11}$ c. $\frac{132}{13}$ d. 132

#### Solution:

1 - (113)111 = 143(1 - (113))143(111)
[Multiply the numerator and denominator by 143, the LCD of 11 and 13.]

= 143-11 / 13 = 132 / 13.
[Simplify.]

Correct answer : (3)
7.
Simplify:
 a. $\frac{43}{24}$ b. $\frac{150}{7}$ c. $\frac{145}{7}$ d. $\frac{25}{4}$

#### Solution:

6 + (14)724

= 24(6 + (14))24(724)
[Multiply numerator and denominator with 24, the LCD of 4 and 24.]

= 144+6 / 7 = 150 / 7.
[Simplify.]

Correct answer : (2)
8.
Find the sum of the infinite geometric series.
1 + $\frac{1}{11}$ + $\frac{1}{121}$ + ......
 a. $\frac{1}{11}$ b. $\frac{10}{11}$ c. $\frac{11}{10}$ d. $\frac{11}{12}$

#### Solution:

r = a2a1 = 1111 = 1 / 11
[The common ratio.]

| r | = | 1 / 11 | = 1 / 11 < 1, so the series converges and the sum exists.

S = a11-r= 11 - (111)
[Use the formula, and replace a1 = 1, r = 1 / 11.]

S = 1(1011) = 11 / 10
[Simplify.]

So, the sum of the infinite geometric series is 11 / 10.

Correct answer : (3)
9.
Find the sum of the infinite geometric series.
1 - $\frac{1}{12}$ + $\frac{1}{144}$ - ......
 a. $\frac{12}{13}$ b. $\frac{12}{11}$ c. $\frac{11}{12}$ d. $\frac{13}{12}$

#### Solution:

r = a2a1 = - 1121 = - 1 / 12
[The common ratio.]

| r | = | - 1 / 12 | = 1 / 12 < 1, so the series converges and the sum exists.

S = a11-r = 11 - (- 112)
[Use the formula, and replace a1 = 1, r = - 1 / 12.]

S = 1(1312) = 12 / 13
[Simplify.]

So, the sum of the infinite geometric series is 12 / 13.

Correct answer : (1)
10.
Find the sum of the infinite geometric series.
- 30 + 1 - $\frac{1}{30}$ + .......
 a. - $\frac{900}{31}$ b. - $\frac{871}{30}$ c. - $\frac{1}{31}$ d. - $\frac{60}{31}$

#### Solution:

r = a2a1 = 1- 30 = - 1 / 30
[Replace a1 = - 30 and a2 = 1.]

| r | = | - 1 / 30 | = 1 / 30 < 1, so the series converges and the sum exists.

S = a11-r = - 301-(- 130)
[Use the formula and replace a1 = - 30, r = - 1 / 30.]

S = - 900 / 31
[Simplify.]

So, the sum of the infinite geometric series is - 900 / 31.

Correct answer : (1)

*AP and SAT are registered trademarks of the College Board.