Any term of a geometric sequence can be found by using the value of the common ratio, the position of the term, and the value of the first term of the sequence. With these values, we can form a formula to find the nth term.

Here, we will look at some solved examples of the nth term of a geometric sequence. In addition, we will see some practice problems.

## Steps to find the nth term of a geometric sequence

Consider the sequence of numbers 2, 6, 18, 54, … Each term in this sequence can be obtained from the previous term by multiplying it by 3. This is an example of a geometric sequence.

Geometric sequences are sequences of numbers in which each term can be obtained by multiplying the previous term by a number called the **common ratio**.

Generally, the terms of a geometric sequence can be obtained using the following formula:

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

where,

- $latex a$ is the first term of the sequence.
- $latex r$ is the common ratio.
- $latex n $ is the position of the term.

Therefore, we can find the nth term of a geometric sequence with the following steps.

#### 1. Identify the value of the first term.

#### 2. Find the value of the common ratio.

The common ratio is found by dividing any term by its previous term.

#### 3. Apply the formula for the nth term.

Use the values of the first term, the common ratio, and the position of the term in the formula $latex a_{n}=ar^{n-1}$.

## 10 Examples of the nth term of geometric sequences with answers

**EXAMPLE 1**

Find the 4th term of a geometric sequence where the first term is 5 and the common ratio is 2.

##### Solution

In this case, we know the values of the first term and the common ratio directly. Then, we can see the following information:

- First term $latex a=5$
- Common ratio: $latex r=2$
- Position of term: $latex n=4$

Now, we apply the formula for the nth term of a geometric sequence with the given values:

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

$latex a_{4}=(5)(2)^{4-1}$

$latex a_{4}=(5)(2)^3$

$latex a_{4}=(5)(8)$

$latex a_{4}=40$

**EXAMPLE **2

**EXAMPLE**

What is the 6th term of a geometric sequence in which the first term is 3 and the common ratio is 3?

##### Solution

As in the previous example, here we also know the values of the first term and the common ratio. Then, we have:

- $latex a=3$
- $latex r=3$
- $latex n=6$

When we use the formula for the nth term of a geometric sequence, we have:

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

$latex a_{6}=(3)(3)^{6-1}$

$latex a_{6}=(3)(3)^5$

$latex a_{6}=(3)(243)$

$latex a_{6}=729$

**EXAMPLE **3

**EXAMPLE**

Find the value of the 6th term of a geometric sequence where the first term is 6 and the common ratio is -2.

##### Solution

We can observe the following values:

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

We see that we have a negative common ratio. However, the formula for the nth term applies in any case:

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

$latex a_{6}=(6)(-2)^{6-1}$

$latex a_{6}=(6)(-2)^5$

$latex a_{6}=(6)(-32)$

$latex a_{6}=-192$

**EXAMPLE **4

**EXAMPLE**

The first four terms of a geometric sequence are 3, 6, 12, and 24. What is the value of the 6th term of the sequence?

##### Solution

In this example, we know the value of the first term, but we don’t know the value of the common ratio of the sequence.

We can find the value of the common ratio by dividing any term by its previous term. For example, $latex \frac{6}{3}=2$. Therefore, we have:

- $latex a=3$
- $latex r=2$
- $latex n=6$

Now, we use these values in the formula for the nth term:

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

$latex a_{6}=(3)(2)^{6-1}$

$latex a_{6}=(3)(2)^5$

$latex a_{6}=(3)(32)$

$latex a_{6}=96$

**EXAMPLE **5

**EXAMPLE**

Find the 6th term of a geometric sequence that starts with the terms -5, 15, -45, …

##### Solution

We have to start by finding the value of the common ratio. Then, we have $latex \frac{15}{-5}=-3$:

- $latex a=-5$
- $latex r=-3$
- $latex n=6$

Now, we can apply the formula for the nth term:

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

$latex a_{6}=(-5)(2)^{6-1}$

$latex a_{6}=(-5)(-3)^5$

$latex a_{6}=(-5)(-243)$

$latex a_{6}=1215$

**EXAMPLE **6

**EXAMPLE**

Find the 10th term of a geometric sequence that starts with the terms 2, 4, 8, …

##### Solution

The common ratio of the progression is $latex \frac{4}{2}=2$. Then, we have the following values:

- $latex a=2$
- $latex r=2$
- $latex n=10$

Now, we can apply the formula for the nth term with these values:

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

$latex a_{10}=(2)(2)^{10-1}$

$latex a_{10}=(2)(2)^9$

$latex a_{10}=(2)(512)$

$latex a_{10}=1024$

**EXAMPLE **7

**EXAMPLE**

If a geometric sequence begins with terms 1, -3, 9, …, what is the value of the 7th term?

##### Solution

The value of the common ratio of the sequence is $latex \frac{-3}{1}=-3$. Then, we have the following:

- $latex a=1$
- $latex r=-3$
- $latex n=7$

Now, we apply the formula for the nth term with the given values:

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

$latex a_{7}=(1)(-3)^{7-1}$

$latex a_{7}=(1)(-3)^6$

$latex a_{7}=(1)(729)$

$latex a_{7}=729$

**EXAMPLE **8

**EXAMPLE**

What is the 9the term of a geometric sequence that starts with the terms 2, 3, $latex 4\frac{1}{2}$?

##### Solution

We find the common ratio by dividing a term by its previous term: $latex \frac{3}{2}$. Therefore, we have:

- $latex a=2$
- $latex r=\frac{3}{2}$
- $latex n=9$

Now, we use the formula for the nth term:

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

$$a_{9}=(2)\left(\frac{3}{2}\right)^{9-1}$$

$$a_{9}=(2)\left(\frac{3}{2}\right)^8$$

$$a_{9}=(2)\left(\frac{6561}{256}\right)$$

$$a_{9}=\frac{6561}{128}$$

$$a_{9}=51\frac{33}{128}$$

**EXAMPLE **9

**EXAMPLE**

The first three terms of a geometric sequence are 81, -54, 36, … What is the value of the 8th term?

##### Solution

The common ratio of the sequence is $latex \frac{-54}{81}=-\frac{2}{3}$. Then, we have the following values:

- $latex a=81$
- $latex r=-\frac{2}{3}$
- $latex n=8$

Applying the formula for the nth term with the given values, we have:

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

$$a_{8}=(81)\left(-\frac{2}{3}\right)^{8-1}$$

$$a_{8}=(81)\left(-\frac{2}{3}\right)^7$$

$$a_{8}=(81)\left(-\frac{128}{2187}\right)$$

$$a_{8}=-\frac{128}{27}$$

$$a_{8}=-4\frac{20}{27}$$

**EXAMPLE **10

**EXAMPLE**

Find the 5th term of a geometric sequence that starts with the terms $latex 2, ~\frac{2}{5}, ~\frac{2}{25}$, …

##### Solution

The common ratio of the sequence is $latex \frac{2}{5}\div 2= \frac{1}{5}$. Therefore, we have the following:

- $latex a=2$
- $latex r=\frac{1}{5}$
- $latex n=5$

Now, we use these values in the formula for the nth term of a geometric sequence:

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

$$a_{5}=(2)\left(\frac{1}{5}\right)^{5-1}$$

$$a_{5}=(2)\left(\frac{1}{5}\right)^4$$

$$a_{5}=(2)\left(\frac{1}{625}\right)$$

$$a_{5}=\frac{2}{625}$$

## nth Term of geometric sequences – Practice problems

#### Find the value of the 100th term of the geometric sequence that begins with the terms 7, -7, 7, …

Write the answer in the input box.

## See also

Interested in learning more about geometric sequences? You can take a look at these pages:

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