Practice Problems on Geometric Series

A geometric series is a type of infinite series formed by summing the terms of a geometric sequence. A geometric sequence is a sequence of numbers where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio.

The general form of a geometric series can be expressed as:

S = a + ar + ar^2 + ar^3 + ar^4 + \ldots

Where:

S is the sum of the series.
a is the first term.
r is the common ratio.

Sample Questions of Geometric Series

Question 1: What is the Geometric mean 2, 4, 8?

Solution:

According to the formula,
=\sqrt [3]{(2)(4)(8)}\\=4

Question 2: Find the first term and common factor in the following Geometric Progression:

4, 8, 16, 32, 64, . . .

Solution:

Here, It is clear that the first term is 4, a=4
We obtain common Ratio by dividing 1st term from 2nd:
r = 8/4 = 2

Question 3: Find the 8^th and the n^thterm for the G.P: 3, 9, 27, 81, . . .

Solution:

Put n=8 for 8^th term in the formula: ar^n-1
For the G.P : 3, 9, 27, 81 . . .
First term (a) = 3
Common Ratio (r) = 9/3 = 3
8^th term = 3(3)^8-1 = 3(3)⁷ = 6561
N^th = 3(3)^n-1 = 3(3)ⁿ(3)^-1
= 3ⁿ

Question 4: For the G.P. : 2, 8, 32, . . . which term will give the value 131073?

Solution:

Assume that the value 131073 is the N^th term,
a = 2, r = 8/2 = 4
N^th term (a_n) = 2(4)^n-1 = 131073
4^n-1 = 131073/2 = 65536
4^n-1 = 65536 = 4⁸
Equating the Powers since the base is same:
n-1 = 8
n = 9

Question 5: Find the sum up to 5^th and N^thterm of the series: 1, \frac{1}{2},\frac{1}{4},\frac{1}{8}...

Solution:

a= 1, r = 1/2
Sum of N terms for the G.P, {S_n =\frac{a(1-r^n)}{1-r}}
= \frac{1(1-(\frac{1}{2})^n)}{1-\frac{1}{2}}
= 2 (1-(\frac{1}{2})^n)
Sum of first 5 terms ⇒ a₅ = 2 ( 1-(\frac{1}{2})^5)
= 2 ( 1-(\frac{1}{32}))
= (\frac{31}{16})

Question 6: Find the Sum of the Infinite G.P: 0.5, 1, 2, 4, 8, ...

Solution:

Formula for the Sum of Infinite G.P: \frac{a}{1-r} ; r≠0
a = 0.5, r = 2
S_∞= (0.5)/(1-2) = 0.5/(-1)= -0.5

Question 7: Find the sum of the Series: 5, 55, 555, 5555,... n terms

Solution:

The given Series is not in G.P but it can easily be converted into a G.P with some simple modifications.
Taking 5 common from the series: 5 (1, 11, 111, 1111,... n terms)
Dividing and Multiplying with 9: \frac{5}{9}(9+ 99+ 999+...n terms)
⇒ \frac{5}{9}[((10+(10)^2+(10)^3+...n terms)-(1+1+1+...n terms)]
⇒\frac{5}{9}[(\frac{10((10)^n-1)}{10-1})-(n)]
⇒ \frac{5}{9}[(\frac{10((10)^n-1)}{9})-(n)]

Worksheet: Geometric Series

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Download Free PDF Worksheet on Geometric Series

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Practice Problems on Geometric Series

Sample Questions of Geometric Series

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