# Sum of Geometric Sequences – Formulas and Examples

Geometric sequences are formed by multiplying each term by the common ratio. To find the sum of these sequences, we need the values of the first term, the common ratio, and the number of terms. Then, we can apply the standard formula for the sum.

Here, we will look at the formulas that we can use to find the sum of a geometric sequence. Then, we will solve some practice problems.

##### ALGEBRA

Relevant for …

Learning to find the sum of a geometric sequence.

See formulas

##### ALGEBRA

Relevant for

Learning to find the sum of a geometric sequence.

See formulas

## Formulas for the sum of a geometric sequence

A geometric sequence is a sequence in which each of its terms is formed by multiplying the previous term by a number called the common ratio.

We can find the sum of the first $latex n$ terms of a geometric sequence using the following formula:

$$S_{n}=a\left( \frac{1-r^n}{1-r}\right)$$

Alternatively, we can write the formula as follows:

$$S_{n}=a\left( \frac{r^n-1}{r-1}\right)$$

where

• $latex a$ is the first term of the sequence.
• $latex r$ is the common ratio.
• $latex n$ is the number of terms.

## Proof of the formula for the sum of geometric sequences

Each term of a geometric sequence is obtained by multiplying the previous term by the common ratio $latex r$. Therefore, we can write the following:

$$S_{n}=a+ar+ar^2+…+ar^{n-1}$$

This is equation [1]. If we multiply both sides of the equation by $latex r$, we have:

$$rS_{n}=ar+ar^2+ar^3+…+ar^n$$

This is equation [2]. If we subtract equation [2] from equation [1], we have:

$$S_{n}-rS_{n}=(a+ar+…+ar^{n-1})-(ar+ar^2+…+ar^n)$$

Simplifying, we have:

$$S_{n}(1-r)=a-ar^n$$

$$S_{n}=\frac{a(1-r^n)}{1-r}$$

We can get the alternate version if we multiply both the numerator and denominator of this formula by -1:

$$S_{n}=\frac{a(r^n-1)}{r-1}$$

## Solved examples of the sum of geometric sequences

### EXAMPLE 1

Find the sum of the first four terms of a geometric sequence, where the first term is equal to 4 and the common ratio is equal to 2.

We start by writing the information we know:

• $latex a=4$
• $latex r=2$
• $latex n=4$

Now, we use the formula for the sum of a geometric sequence:

$$S_{n}=\frac{a(r^n-1)}{r-1}$$

$$S_{4}=\frac{4(2^4-1)}{2-1}$$

$$=\frac{4(16-1)}{1}$$

$$=\frac{4(15)}{1}$$

$$S_{4}=60$$

### EXAMPLE 2

What is the sum of the first 5 terms of a geometric sequence in which the first term is equal to -5 and the common ratio is -2?

We have the following information:

• $latex a=-5$
• $latex r=-2$
• $latex n=5$

Applying the formula for the sum of geometric sequences, we have:

$$S_{n}=\frac{a(r^n-1)}{r-1}$$

$$S_{5}=\frac{-5((-2)^5-1)}{-2-1}$$

$$=\frac{-5(-32-1)}{-3}$$

$$=\frac{-5(-33)}{-3}$$

$$=\frac{165}{-3}$$

$$S_{5}=-55$$

### EXAMPLE 3

A geometric sequence has a first term equal to 1 and a common ratio equal to $latex \frac{1}{4}$. Find the sum of the first four terms.

We have the following information:

• $latex a=1$
• $latex r=\frac{1}{4}$
• $latex n=4$

When we use the formula for the sum of geometric sequences, we have:

$$S_{n}=\frac{a(1-r^n)}{1-r}$$

$$S_{4}=\frac{1(1-(\frac{1}{4})^4)}{1-\frac{1}{4}}$$

We can factor the numerator using the difference of squares twice:

$$S_{4}=\frac{(1-(\frac{1}{4})^2)(1+(\frac{1}{4})^2)}{1-\frac{1}{4}}$$

$$=\frac{(1-\frac{1}{4})(1+\frac{1}{4})(1+(\frac{1}{4})^2)}{1-\frac{1}{4}}$$

$$=\left(1+\frac{1}{4}\right)\left(1+(\frac{1}{4})^2\right)$$

$$=\left(\frac{5}{4}\right)\left(\frac{17}{16}\right)$$

$$S_{4}=\frac{85}{64}$$

### EXAMPLE 4

Find the sum of the first 10 terms of the geometric sequence starting with 3, 6, 12, …

In this case, we don’t know the value of the common ratio directly, but we just have to divide a term by its previous term to find it: 6/3=2. Then, we have:

• $latex a=3$
• $latex r=2$
• $latex n=10$

Using the formula for the sum with the given values, we have:

$$S_{n}=\frac{a(r^n-1)}{r-1}$$

$$S_{10}=\frac{3(2^{10}-1)}{2-1}$$

$$=\frac{3(1024-1)}{1}$$

$latex =3(1023)$

$latex S_{10}=3069$

### EXAMPLE 5

Find the sum of the first 6 terms of the geometric sequence that starts with the terms -2, 8, -32, …

Similar to the previous example, we start by finding the common ratio. Then, we have 8/-2=-4. Therefore, we have the following:

• $latex a=-2$
• $latex r=-4$
• $latex n=6$

Applying the formula for the sum of a geometric sequence, we have:

$$S_{n}=\frac{a(r^n-1)}{r-1}$$

$$S_{6}=\frac{-2((-4)^6-1)}{-4-1}$$

$$=\frac{-2(4096-1)}{-5}$$

$$=\frac{-2(4095)}{-5}$$

$latex S_{6}=1638$

### EXAMPLE 6

Find the sum of the following geometric sequence:

$$3+6+12+…+384$$

In this case, we know neither the common ratio nor the number of terms. Therefore, we start by finding the common ratio: 6/3=2.

To find the number of terms, we can use the formula for the nth term of a geometric sequence, $latex a_{n}=ar^{n-1}$ and solve for $latex n$:

$latex a_{n}=ar^{n-1}$

$latex 384=(3)(2)^{n-1}$

$latex 128=2^{n-1}$

$$128=\frac{2^n}{2^1}$$

$latex 256=2^n$

$latex n=\log_{2}(256)$

$latex n=8$

Now, we use the formula for the sum with the found values:

$$S_{n}=\frac{a(r^n-1)}{r-1}$$

$$S_{8}=\frac{3(2^8-1)}{2-1}$$

$$=\frac{3(256-1)}{1}$$

$latex =3(255)$

$latex S_{8}=765$

You can explore additional examples on this topic in our article: Sum of Geometric Sequences – Examples and Practice Problems

## Sum of geometric sequences – Practice problems

Sum of geometric sequences quiz
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#### Find the sum of the following geometric sequence $$7-14+28-…+448$$

Write the answer in the input box.

$latex S=$