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 \[ 2 , 8 , 32 , 128 , ... \] is obtained starting from \( 2 \) and multiplying each term by \( 4 \). \( 2 \) is the first term of the sequence and \( 4 \) is the common ratio. \[ \begin{aligned} 8 &= 2 \times 4 \\ 32 &= 8 \times 4 \\ 128 &= 32 \times 4 \\ &\text{and so on} \end{aligned} \]
The terms in the sequence may also be written as follows \begin{align*} 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 \end{align*}
The \( n^{ th} \) term may now be written as \[ a_n = a_1 r^{n-1} \]
where a₁ 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 \( |r| \lt 1 \) 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.

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 \begin{align*} a_2 &= a_1 \times r = 10 \cdot (-1) = -10 \\ a_3 &= a_2 \times r = -10 \cdot (-1) = 10 \\ a_4 &= a_3 \times r = 10 \cdot (-1) = -10 \\ a_5 &= a_4 \times r = -10 \cdot (-1) = 10 \end{align*}

Problem 2

Find the 10th term of a geometric sequence if \(a_1 = 45\) and the common ratio \(r = 0.2\).

Solution to Problem 2:

Use the formula \[ a_n = a_1 \times r^{\,n-1} \] that gives the \(n\)th term to find \(a_{10}\) 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 \[ -\frac{1}{2}, \frac{1}{4}, -\frac{1}{8}, \frac{1}{16}, \dots \]

Solution to Problem 3:

We first use the first few terms to find the common ratio \(r\): \[ r = \frac{a_2}{a_1} = \frac{\frac{1}{4}}{-\frac{1}{2}} = -\frac{1}{2}, \\ r = \frac{a_3}{a_2} = \frac{-\frac{1}{8}}{\frac{1}{4}} = -\frac{1}{2} \] The common ratio is \(r = -\frac{1}{2}\). We now use the formula \(a_n = a_1 r^{\,n-1}\) for the \(n\)th term to find \(a_{20}\) as follows: \[ a_{20} = a_1 \times r^{20-1} = \left(-\frac{1}{2}\right) \times \left(-\frac{1}{2}\right)^{19} = \left(-\frac{1}{2}\right)^{20} = \frac{1}{2^{20}} \]

Problem 4

Given the terms \(a_{10} = \frac{3}{512}\) and \(a_{15} = \frac{3}{16384}\) of a geometric sequence, find the exact value of the term \(a_{30}\) of the sequence.

Solution to Problem 4:

We first use the formula for the \(n\)th term to write \(a_{10}\) and \(a_{15}\) as follows:
\[ a_{10} = a_1 \times r^{10-1} = a_1 \; r^9 = \frac{3}{512} \] \[ a_{15} = a_1 \times r^{15-1} = a_1 \; r^{14} = \frac{3}{16384} \]
We now divide the terms \(a_{15}\) and \(a_{10}\) to write:
\[ \frac{a_{15}}{a_{10}} = \frac{a_1 r^{14}}{a_1 r^9} = \frac{3 / 16384}{3 / 512} \] Simplify expressions in the above equation to obtain: \[ r^5 = \frac{1}{32} \quad \text{which gives} \quad r = \frac{1}{2} \] We now use \(a_{10}\) to find \(a_1\) as follows: \[ a_{10} = \frac{3}{512} = a_1 \left(\frac{1}{2}\right)^9 \] 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: \[ a_{30} = 3 \left(\frac{1}{2}\right)^{29} = \frac{3}{536870912} \]

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 note that the terms making the sum are those of a geometric sequence with \(a_1 = 1\) and \(r = 3\). Using the formula \[ S_n = a_1 \frac{1 - r^n}{1 - r} \quad \text{with } n = 6 \] we get: \[ S_6 = 1 \cdot \frac{1 - 3^6}{1 - 3} = 364 \]

Problem 6

Find the sum \[ S = \sum_{i=1}^{10} 8 \times \left(\frac{1}{4}\right)^{i - 1} \]

Solution to Problem 6:

An examination of the terms included in the sum are
\[ 8, \; 8 \times \left(\frac{1}{4}\right)^1, \; 8 \times \left(\frac{1}{4}\right)^2, \; \dots, \; 8 \times \left(\frac{1}{4}\right)^9 \]
These are the terms of a geometric sequence with \(a_1 = 8\) and \(r = \frac{1}{4}\), and therefore we can use the formula for the sum of the first \(n\) terms of a geometric sequence:
\[ S_{10} = a_1 \frac{1 - r^{10}}{1 - r} \]
\[ S_{10} = 8 \frac{1 - \left(\frac{1}{4}\right)^{10}}{1 - \frac{1}{4}} \approx 10.67 \quad \text{(rounded to 2 decimal places)} \]

Problem 7

Write the rational number \(5.31313131\ldots\) as the ratio of two integers.

Solution to Problem 7:

We first write the given rational number as an infinite sum as follows: \[ 5.313131\ldots = 5 + 0.31 + 0.0031 + 0.000031 + \ldots \] The terms \(0.31 + 0.0031 + 0.000031 + \ldots\) are those of a geometric sequence with \(a_1 = 0.31\) and \(r = 0.01\). Hence, we can use the formula for the sum of an infinite geometric sequence: \[ S = \frac{a_1}{1 - r} = \frac{0.31}{1 - 0.01} = \frac{0.31}{0.99} = \frac{31}{99} \] We now write \(5.313131\ldots\) as follows: \[ 5.313131\ldots = 5 + \frac{31}{99} = \frac{526}{99} \]

Exercises with Answers

Answer the following questions related to geometric sequences:
a) Find \(a_{20}\) given that \(a_3 = \frac{1}{2}\) and \(a_5 = 8\)
b) Find \(a_{30}\) given that the first few terms of a geometric sequence are given by \(-2, 1, -\frac{1}{2}, \frac{1}{4}, \ldots\)
c) Find \(r\) given that \(a_1 = 10\) and \(a_{20} = 10^{-18}\)
d) Write the rational number \(0.9717171\ldots\) as a ratio of two positive integers.

Answers

a) \(a_{20} = 2^{18}\)
b) \(a_{30} = \frac{1}{2^{28}}\)
c) \(r = 0.1\)
d) \(0.9717171\ldots = \frac{481}{495}\)