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

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.

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

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

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

### General Term

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

$a_n = a_1 \cdot r^{n - 1}$

### Sum of Terms

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

$S_n = \frac{a (1 - r^n)}{1 - r}, \hspace{0.75em} r \neq 1$

#### A Special Case

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

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

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

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

### Examples

Let's look at a couple of simple examples.

Example

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

Solution

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

$a_n \hspace{0.25em} = \hspace{0.25em} a_1 \cdot r^{n - 1}$

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

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

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

Example

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

Solution

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

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

$a_n \hspace{0.25em} = \hspace{0.25em} a_1 \cdot r^{n - 1}$

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

$\begin{align*} a_5 \hspace{0.25em} &= \hspace{0.25em} a_1 \cdot 2^{5 - 1} \\[1em] 162 \hspace{0.25em} &= \hspace{0.25em} a_1 \cdot r^4 \hspace{0.25cm} \rule[0.1cm]{1cm}{0.1em} \hspace{0.15cm} (1) \end{align*}$

$\begin{align*} a_8 \hspace{0.25em} &= \hspace{0.25em} a_1 \cdot 2^{8 - 1} \\[1em] 6 \hspace{0.25em} &= \hspace{0.25em} a_1 \cdot r^7 \hspace{0.25cm} \rule[0.1cm]{1cm}{0.1em} \hspace{0.15cm} (2) \end{align*}$

Dividing the second equation by the first, we have

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

Taking the cube root on both sides,

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

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

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

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

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

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

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

Done.