# Geometric Sequences Problems with Solutions

Geometric sequences are used in several branches of applied mathematics to engineering, sciences, computer sciences, biology, finance...
Problems and exercises involving geometric sequences, along with answers are presented.

## Review OF Geometric Sequences

The sequence shown below

2 , 8 , 32 , 128 , ...
has been obtained starting from 2 and multiplying each term by 4. 2 is the first term of the sequence and 4 is the common ratio . $8 = 2 \times 4 \\ 32 = 8 \times 4 \\ 128 = 32 \times 4 \\ \text{and so on}$
The terms in the sequence may also be written as follows
$a_1 = 2 \\ a_2 = a_1 \times 4 = 2 \times 4 \\ a_3 = a_2 \times 4 = 2 \times 4^2 \\ a_4 = a_3 \times 4 = 2 \times 4^3 \\$
The n th term may now be written as $a_n = a_1 r^{n-1}$
where a
1 is the first term of the sequence and r is the common ratio which is equal to 4 in the above example.
The sum of the first n terms of a geometric sequence is given by $s_n = a_1 + a_2 + a_3 + ... a_n = a_1 \dfrac{1 - r^n}{1-r}$ The sum S of an infinite (n approaches infinity) geometric sequence and when |r| < 1 is given by

$S = \dfrac{a_1}{1-r}$
Arithmetic Series Online Calculator . An online calculator to calculate the sum of the terms in an arithmetic sequence.

## Problems with Solutions

Problem 1
Find the terms a2, a3, a4 and a5 of a geometric sequence if a1 = 10 and the common ratio r = - 1.

Solution to Problem 1:
Use the definition of a geometric sequence $a_2 = a_1 \times r = 10 (-1) = - 10 \\ a_3 = a_2 \times r = - 10 (-1) = 10 \\ a_4 = a_3 \times r = 10 (-1) = - 10 \\ a_5 = a_4 \times r = - 10 (-1) = 10$

Problem 2

Find the 10 th term of a geometric sequence if a1 = 45 and the common ration r = 0.2.

Solution to Problem 2:
Use the formula $a_n = a_1 \times r^{n-1}$ that gives the n th term to find a10 as follows
$a_{10} = 45 \times 0.2^{10-1} = 2.304 \times 10^{-5}$

Problem 3

Find a20 of a geometric sequence if the first few terms of the sequence are given by

-1/2 , 1/4 , -1/8 , 1 / 16 , ...

Solution to Problem 3:
We first use the first few terms to find the common ratio r
$r = a_2 / a_1 = (1/4) / (-1/2) = -1/2 \\ \\ r = a_3 / a_2 = (-1/8) / (1/4) = -1/2 \\$
The common ration r = -1/2. We now use the formula an = a1 r n-1 for the n th term to find a20 as follows.
$a_20 = a_1 \times r^{20-1} = (-1/2)\times (-1/2)^{19} = (-1/2)^{20} = 1 / 2^{20}$

Problem 4

Given the terms a10 = 3 / 512 and a15 = 3 / 16384 of a geometric sequence, find the exact value of the term a30 of the sequence.

Solution to Problem 4:
We first use the formula for the n th term to write a10 and a15 as follows
$a_{10} = a_1 \times r^{10-1} = a_1 r^9 = 3 / 512 \\ \\ a_{15} = a_1 \times r^{15-1} = a_1 r^{14} = 3 / 16384$
We now divide the terms a10 and a15 to write
$a_{15} / a_{10} = a_1 \times r^{14} / (a_1 \times r^9) = (3 / 16384) / (3 / 512)$
Simplify expressions in the above equation to obtain.
r5 = 1 / 32 which gives r = 1/2
We now use a10 to find a1 as follows.
$a_{10} = 3 / 512 = a_1 (1/2)^9$ Solve for a1 to obtain.
$a_1 = 3$
We now use the formula for the n th term to find a30 as follows.
$a_{30} = 3(1/2)^{29} = 3 / 536870912$

Problem 5

Find the sum
$S = \sum_{k=1}^{6} 3^{k - 1}$

Solution to Problem 5:
We first rewrite the sum S as follows
S = 1 + 3 + 9 + 27 + 81 + 243 = 364
Another method is to first note that the terms making the sum are those of a geometric sequence with a1 = 1 and r = 3 using the formula sn = a1 (1 - rn) / (1 - r) with n = 6.
s6 = 1 (1 - 36) / (1 - 3) = 364

Problem 6

Find the sum
$S = \sum_{i=1}^{10} 8 \times (1/4)^{i - 1}$

Solution to Problem 6:
An examination of the terms included in the sum are
8 , 8× ((1/4)1 , 8×((1/4)2 , ... , 8×((1/4)9
These are the terms of a geometric sequence with a1 = 8 and r = 1/4 and therefore we can use the formula for the sum of the terms of a geometric sequence
s10 = a1 (1 - rn) / (1 - r)
= 8 × (1 - (1/4)10) / (1 - 1/4) = 10.67 (rounded to 2 decimal places)

Problem 7

Write the rational number 5.31313131... as the ratio of two integers.

Solution to Problem 7:
We first write the given rational number as an infinite sum as follows
5.313131... = 5 + 0.31 + 0.0031 + 0.000031 + ....
The terms making 0.31 + 0.0031 + 0.000031 ... are those of a geometric sequence with a1 = 0.31 and r = 0.01. Hence the use of the formula for an infinite sum of a geometric sequence
S = a1 / (1 - r) = 0.31 / (1 - 0.01) = 0.31 / 0.99 = 31 / 99
We now write 5.313131... as follows
5.313131... = 5 + 31/99 = 526 / 99

Answer the following questions related to geometric sequences:
a) Find a20 given that a3 = 1/2 and a5 = 8
b) Find a30 given that the first few terms of a geometric sequence are given by -2 , 1 , -1/2 , 1/4 ...
c) Find r given that a1 = 10 and a20 = 10-18
d) write the rational number 0.9717171... as a ratio of two positive integers.