# 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 detailed solutions and answers, are presented.

__REVIEW OF GEOMETRIC SEQUENCES__

The sequence shown below

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}

S = \dfrac{a_1}{1-r}

Arithmetic Series Online Calculator . An online calculator to calculate the sum of the terms in an arithmetic sequence.

__Problem 1__

Find the terms a_{2}, a_{3}, a_{4} and a_{5} of a geometric sequence if a_{1} = 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 a

_{1}= 45 and the common ration r = 0.2.

**Use the formula**

__Solution to Problem 2:__
a_n = a_1 \times r^{n-1}

_{10}as follows

a_{10} = 45 \times 0.2^{10-1} = 2.304 \times 10^{-5}

__Problem 3__

Find a

_{20}of a geometric sequence if the first few terms of the sequence are given by

__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 a_{n}= a

_{1}r

^{ n-1}for the n th term to find a

_{20}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 a

_{10}= 3 / 512 and a

_{15}= 3 / 16384 of a geometric sequence, find the exact value of the term a

_{30}of the sequence.

**We first use the formula for the n th term to write a**

__Solution to Problem 4:___{10}and a

_{15}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 a_{10}and a

_{15}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. r

^{5}= 1 / 32 which gives r = 1/2

We now use a

_{10}to find a

_{1}as follows.

a_{10} = 3 / 512 = a_1 (1/2)^9

Solve for a_{1}to obtain.

a_1 = 3

We now use the formula for the n th term to find a_{30}as follows.

a_{30} = 3(1/2)^{29} = 3 / 536870912

__Problem 5__

Find the sum

S = \sum_{k=1}^{6} 3^{k - 1}

**We first rewrite the sum S as follows**

__Solution to Problem 5:__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 a

_{1}= 1 and r = 3 using the formula s

_{n}= a

_{1}(1 - r

^{n}) / (1 - r) with n = 6.

s

_{6}= 1 (1 - 3

^{6}) / (1 - 3) = 364

__Problem 6__

Find the sum

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

**An examination of the terms included in the sum are**

__Solution to Problem 6:__8 , 8× ((1/4)

^{1}, 8×((1/4)

^{2}, ... , 8×((1/4)

^{9}

These are the terms of a geometric sequence with a

_{1}= 8 and r = 1/4 and therefore we can use the formula for the sum of the terms of a geometric sequence

s

_{10}= a

_{1}(1 - r

^{n}) / (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.

**We first write the given rational number as an infinite sum as follows**

__Solution to Problem 7:__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 a

_{1}= 0.31 and r = 0.01. Hence the use of the formula for an infinite sum of a geometric sequence

S = a

_{1}/ (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

**Exercises:**

Answer the following questions related to geometric sequences:

a) Find a

_{20}given that a

_{3}= 1/2 and a

_{5}= 8

b) Find a

_{30}given that the first few terms of a geometric sequence are given by -2 , 1 , -1/2 , 1/4 ...

c) Find r given that a

_{1}= 10 and a

_{20}= 10

^{-18}

d) write the rational number 0.9717171... as a ratio of two positive integers.

**Answers to Above Exercises:**

a) a

_{20}= 2

^{18}

b) a

_{30}= 1 / 2

^{28}

c) r = 0.1

d) 0.9717171... = 481/495

More math problems with detailed solutions in this site.

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Updated: 28 July 2018 (A Dendane)