Geometric Sequence Calculator

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About the Geometric Sequence Calculator

The geometric sequence calculator lets you calculate various important values for an geometric sequence. You can calculate the first term, nth\hspace{0.2em} n^{\text{th}} \hspace{0.2em}term, common ratio, sum of n\hspace{0.2em} n \hspace{0.2em} terms, number of terms, or position of a term in the geometric sequence.

The calculator will not only give you the answer but also a step-by-step solution.

Usage Guide

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i. Valid Inputs

The different inputs (except n\hspace{0.2em} n \hspace{0.2em}) can be a real number in any format as shown below.

  • Whole numbers or decimals → 2\hspace{0.2em} 2 \hspace{0.2em}, 4.25\hspace{0.2em} -4.25 \hspace{0.2em}, 0\hspace{0.2em} 0 \hspace{0.2em}, 0.33\hspace{0.2em} 0.33 \hspace{0.2em}
  • Fractions → 2/3\hspace{0.2em} 2/3 \hspace{0.2em}, 1/5\hspace{0.2em} -1/5 \hspace{0.2em}
  • Mixed numbers → 51/4\hspace{0.2em} 5 \hspace{0.5em} 1/4 \hspace{0.2em}

Number of terms n\hspace{0.2em} n \hspace{0.2em} must be a positive integer (a counting number).

Note — Because of the exponential nature of the sequence, the terms may become quickly too large or too small to be calculated or displayed accurately by the caluclator.

ii. Example

If you would like to see an example of the calculator's working, just click the "example" button.

iii. Solutions

As mentioned earlier, the calculator won't just tell you the answer but also the steps you can follow to do the calculation yourself. The "show/hide solution" button would be available to you after the calculator has processed your input.

iv. Share

We would love to see you share our calculators with your family, friends, or anyone else who might find it useful.

By checking the "include calculation" checkbox, you can share your calculation as well.

Here's a quick overview of what a geometric sequence is.

Geometric Sequence

A geometric sequence is a sequence of numbers such that the ratio between any two consecutive terms is constant.

Here's an example of an geometric sequence.

2,6,18,54,162,486,...2,\hspace{0.35em} 6,\hspace{0.35em} 18,\hspace{0.35em} 54,\hspace{0.35em} 162,\hspace{0.35em} 486, ...

The constant ratio is known as the common ratio and is usually denoted by r\hspace{0.2em} r \hspace{0.2em}. For the sequence above, the common ratio would be 6/2=3\hspace{0.2em} 6 / 2 = 3 \hspace{0.2em}.

The nth\hspace{0.2em} n^{\text{th}} \hspace{0.2em} term of an geometric sequence is denoted by an\hspace{0.2em} a_n \hspace{0.2em}.

General Term

For a geometric sequence with a first term a1\hspace{0.2em} a_1 \hspace{0.2em} and common ratio r\hspace{0.2em} r \hspace{0.2em}, the nth\hspace{0.2em} n^{\text{th}} \hspace{0.2em} term is given by —

an=a1rn1a_n = a_1 \cdot r^{n - 1}

Sum of Terms

The sum of n\hspace{0.2em} n \hspace{0.2em} terms (denoted by Sn\hspace{0.2em} S_n \hspace{0.2em}) of a geometric sequence would be

Sn=a(1rn)1r,r1S_n = \frac{a (1 - r^n)}{1 - r}, \hspace{0.75em} r \neq 1

A Special Case

Consider a geometric sequence with a common difference is between 1\hspace{0.2em} -1 \hspace{0.2em} and 1\hspace{0.2em} 1 \hspace{0.2em}. So r<1\hspace{0.2em} | \hspace{0.15em} r \hspace{0.15em}| < 1 \hspace{0.2em}.

Now, as n\hspace{0.2em} n \hspace{0.2em} becomes larger, rn\hspace{0.2em} r^n \hspace{0.2em} becomes smaller. As n\hspace{0.2em} n \hspace{0.2em} approaches infinity rn\hspace{0.2em} r^n \hspace{0.2em} approaches 0\hspace{0.2em} 0 \hspace{0.2em}.

Putting it all together, for an infinite geometric sequence with r<0\hspace{0.2em} | \hspace{0.15em} r \hspace{0.15em}| < 0 \hspace{0.2em} , the sum of terms S\hspace{0.2em} S \hspace{0.2em} is

S=a(1rn)1r=a(10)1r=a1r\begin{align*} S &= \frac{a (1 - {\color{Red} r^n} )}{1 - r} \\[1.75em] &= \frac{a (1 - {\color{Red} 0} )}{1 - r} \\[1.75em] &= \frac{a}{1 - r} \end{align*}


Let's look at a couple of simple examples.


Given the first term a1=4\hspace{0.2em} a_1 = 4 \hspace{0.2em} and the common ratio r=2\hspace{0.2em} r = 2 \hspace{0.2em}, find the 7th\hspace{0.2em} 7^{\text{th}} \hspace{0.2em} term of the geometric sequence.


We know the nth\hspace{0.2em} n^{\text{th}} \hspace{0.2em} term of a geometric progression is given by —

an=a1rn1a_n \hspace{0.25em} = \hspace{0.25em} a_1 \cdot r^{n - 1}

Substituting the values of a1\hspace{0.2em} a_1 \hspace{0.2em}, r\hspace{0.2em} r \hspace{0.2em}, and n\hspace{0.2em} n \hspace{0.2em}, we have

a7=4271=256\begin{align*} a_7 \hspace{0.25em} &= \hspace{0.25em} 4 \cdot 2^{7 - 1} \\[1em] &= \hspace{0.25em} 256 \end{align*}

So the 7th\hspace{0.2em} 7^{\text{th}} \hspace{0.2em} term of the geometric sequence is 256\hspace{0.2em} 256 \hspace{0.2em}.


The 5th\hspace{0.2em} 5^{\text{th}} \hspace{0.2em} term of a geometric sequence is 162\hspace{0.2em} 162 \hspace{0.2em} and the 8th\hspace{0.2em} 8^{\text{th}} \hspace{0.2em} is 6\hspace{0.2em} 6 \hspace{0.2em}. Find the sum of this infinite geometric sequence.


The first thing to note is that the sum of an infinite geometric sequence exists only if the common ratio is smaller than 1\hspace{0.2em} 1 \hspace{0.2em} in absolute value. So, let's calculate r\hspace{0.2em} r \hspace{0.2em} and see whether this condition is satisfied.

Here's how we can calculate r\hspace{0.2em} r \hspace{0.2em}. We know

an=a1rn1a_n \hspace{0.25em} = \hspace{0.25em} a_1 \cdot r^{n - 1}

The question gives us the values of the 5th\hspace{0.2em} 5^{\text{th}} \hspace{0.2em} and the 8th\hspace{0.2em} 8^{\text{th}} \hspace{0.2em} terms. We can use them to get two equations.

a5=a1251162=a1r4(1)\begin{align*} a_5 \hspace{0.25em} &= \hspace{0.25em} a_1 \cdot 2^{5 - 1} \\[1em] 162 \hspace{0.25em} &= \hspace{0.25em} a_1 \cdot r^4 \hspace{0.25cm} \rule[0.1cm]{1cm}{0.1em} \hspace{0.15cm} (1) \end{align*}
a8=a12816=a1r7(2)\begin{align*} a_8 \hspace{0.25em} &= \hspace{0.25em} a_1 \cdot 2^{8 - 1} \\[1em] 6 \hspace{0.25em} &= \hspace{0.25em} a_1 \cdot r^7 \hspace{0.25cm} \rule[0.1cm]{1cm}{0.1em} \hspace{0.15cm} (2) \end{align*}

Dividing the second equation by the first, we have

6162=a1r7a1r4127=r3\begin{align*} \frac{6}{162} \hspace{0.25em} &= \hspace{0.25em} \frac{a_1 \cdot r^7}{a_1 \cdot r^4} \\[1.75em] \frac{1}{27} \hspace{0.25em} &= \hspace{0.25em} r^{3} \end{align*}

Taking the cube root on both sides,

r=13r \hspace{0.25em} = \hspace{0.25em} \frac{1}{3}

Great, since r<1\hspace{0.2em} r < 1 \hspace{0.2em}, the sum of the infinite sequence exists. And as we saw earlier, it is given by the formula —

S=a11rS \hspace{0.25em} = \hspace{0.25em} \frac{a_1}{1 - r}

Now, to be able to use this formula, we need to find a1\hspace{0.2em} a_1 \hspace{0.2em} too. For that, we can use the equation (1)\hspace{0.2em} (1) \hspace{0.2em}.

162=a1r4162=a1(13)4a1=13122\begin{align*} 162 \hspace{0.25em} &= \hspace{0.25em} a_1 \cdot r^4 \\[1.75em] 162 \hspace{0.25em} &= \hspace{0.25em} a_1 \cdot \left ( \frac{1}{3} \right )^4 \\[1.75em] a_1 \hspace{0.25em} &= \hspace{0.25em} 13122 \end{align*}

Finally, substituting the values of a1\hspace{0.2em} a_1 \hspace{0.2em} and r\hspace{0.2em} r \hspace{0.2em} in the formula for sum, we get

S=a11r=13122113=19683\begin{align*} S \hspace{0.25em} &= \hspace{0.25em} \frac{a_1}{1 - r} \\[1.75em] &= \hspace{0.25em} \frac{13122}{1 - \frac{1}{3}} \\[1.75em] &= \hspace{0.25em} 19683 \end{align*}


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