Geometric sequences have the main characteristic of having a common ratio, which is multiplied by the last term to find the next term. Any term in a geometric sequence can be found using a formula.

Here, we will look at a summary of geometric sequences and we will explore its formula. In addition, we will see several examples with answers and exercises to solve to practice these concepts.

## Summary of geometric sequences

Geometric sequences are sequences in which the next number in the sequence is found by multiplying the previous term by a number called the **common ratio**. The common ratio is denoted by the letter *r*.

Depending on the common ratio, the geometric sequence can be increasing or decreasing. If the common ratio is greater than 1, the sequence is increasing and if the common ratio is between 0 and 1, the sequence is decreasing:

We can find any number in the geometric sequence using the geometric sequence formula:

We can find the common ratio by dividing any term by the previous term:

$latex r=\frac{a_{n}}{a_{n-1}}$

## Geometric sequences – Examples with answers

**EXAMPLE 1**

Find the next term in the geometric sequence: 4, 8, 16, 32, **?**.

##### Solution

First, we have to find the common ratio of the geometric progression. To do this, we divide a term by the previous term:

- $latex \frac{32}{16}=2$
- $latex \frac{16}{8}=2$
- $latex \frac{8}{4}=2$

Therefore, the common ratio is 2. To find the next term, we multiply the last term by the common ratio: $latex 32\times 2=64$.

**EXAMPLE 2**

What is the next term in the geometric sequence? 3, 15, 75, 375, **?**.

##### Solution

We start by finding the common ratio for the geometric progression. Then, we divide each term by its previous term:

- $latex \frac{375}{75}=5$
- $latex \frac{75}{15}=5$
- $latex \frac{15}{3}=5$

We see that the common ratio is 5. We find the next term by multiplying the last term by the common ratio : $latex 375 \times 5=1875$.

**EXAMPLE 3**

Determine the next term in the geometric sequence: 48, 24, 12, 6, **?**.

##### Solution

Again, we start by finding the common ratio in the progression:

- $latex \frac{6}{12}=0.5$
- $latex \frac{12}{24}=0.5$
- $latex \frac{24}{48}=0.5$

In this case, we see that the common ratio is between 0 and 1, so the progression is slowing down. The next term in the geometric progression is $latex 6\times 0.5=3$.

**EXAMPLE **4

What is the value of the 6th term of a geometric sequence where the first term is 3 and the common ratio is 2?

##### Solution

We have the following values:

- First term: $latex a_{1}=3$
- Common ratio: $latex r=2$
- Position of term: $latex n=6$

Then, we can use the formula for geometric sequences with the given values:

$latex a_{n}=a_{1}(r^{n-1})$

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

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

$latex a_{6}=5(32)$

$latex a_{6}=160$

**EXAMPLE **5

Find the 12th term in the geometric sequence: 5, 15, 45, 135, …

##### Solution

In this case, we have to use the formula of geometric progressions $latex a_{n}=a_{1}({{r}^{n-1}})$. Therefore, we have to identify the first term, the common reason and the position of the term:

- First term: $latex a_{1}=5$
- Common ratio: $latex r=3$
- Position of term: $latex n=12$

Now, we substitute this data into the formula:

$latex a_{n}=a_{1}({{r}^{n-1}})$

$latex a_{12}=5({{3}^{12-1}})$

$latex a_{12}=5({{3}^{11}})$

$latex a_{12}=5(177147)$

$latex a_{12}=885 735$

We see that we have a very large number. Geometric progressions tend to grow rapidly depending on the common proportion.

**EXAMPLE **6

Find the 8th term in the geometric sequence 8, 32, 128, 512, …

##### Solution

Again, we start by identifying the first term, the common ratio, and the position of the term to be used with the formula:

- First term: $latex a_{1}=8$
- Common ratio: $latex r=4$
- Position of term: $latex n=8$

Now, we use the formula with these values:

$latex a_{n}=a_{1}({{r}^{n-1}})$

$latex a_{8}=8({{4}^{8-1}})$

$latex a_{8}=8({{4}^{7}})$

$latex a_{8}=8(16384)$

$latex a_{8}=131072$

**EXAMPLE **7

Find the 10th term in the geometric sequence: 168, 84, 42, 21, …

##### Solution

In this case, we have a decreasing geometric progression, so we expect the common ratio to be between 0 and 1:

- First term: $latex a_{1}=168$
- Common ratio: $latex r=0.5$
- Position of term: $latex n=10$

We use the formula to find the term 10:

$latex a_{n}=a_{1}({{r}^{n-1}})$

$latex a_{10}=168({{0.5}^{10-1}})$

$latex a_{10}=168({{0.5}^{9}})$

$latex a_{10}=168(0.001953)$

$latex a_{10}=0.328$

**EXAMPLE **8

Find the 7th term in the geometric sequence: 540, 180, 60, 20, …

##### Solution

Similar to the previous example, here we have a decreasing geometric progression, so the common ratio must be between 0 and 1:

- First term: $latex a_{1}=540$
- Common ratio: $latex r=\frac{1}{3}$
- Possiion of term: $latex n=7$

We use these values to substitute in the formula:

$latex a_{n}=a_{1}({{r}^{n-1}})$

$latex a_{7}=540({{\left( \frac{1}{3}\right)}^{7-1}})$

$latex a_{7}=540({{\left( \frac{1}{3}\right)}^{6}})$

$latex a_{7}=540(0.0013717)$

$latex a_{7}=0.7407$

**EXAMPLE **9

**EXAMPLE**

If the 4th term of a geometric sequence is 16 and the 7th term is 128, what is the 11th term?

##### Solution

In this case, we know neither the value of the first term nor the common ratio. However, we can start by forming the following equations:

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

$latex 16=a_{1}(r^{3})~~~[1]$

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

$latex 128=a_{1}(r^{6})~~~[2]$

If we divide equation 2 by equation 1, we have:

$$\frac{128}{16}=\frac{a_{1}(r^{6})}{a_{1}(r^{3})$$

$latex 8=r^{3}$

$latex r=2$

If we consider the 7th term as the 1st term, the 11th term is now the 5th term:

- First term: $latex a_{1}=128$
- Common ratio: $latex r=2$
- Position of term: $latex n=5$

Using these values in the formula, we have:

$latex a_{n}=a_{1}(r^{n-1})$

$latex a_{5}=128(2^{5-1})$

$latex a_{5}=128(2^4)$

$latex a_{5}=128(16)$

$latex a_{5}=2048$

Then, the 11th term of the given sequence is 2048.

**EXAMPLE **10

**EXAMPLE**

A geometric sequence has a 3rd term equal to 256 and an 8th term equal to -8. What is the value of the 14th term?

##### Solution

Similar to the previous example, we can find the common ratio by forming the following equations:

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

$latex 256=a_{1}(r^{2})~~~[1]$

$latex a_{8}=a_{1}(r^{8-1})$

$latex -8=a_{1}(r^{7})~~~[2]$

Now we divide them to obtain:

$$\frac{-8}{256}=\frac{a_{1}(r^{7})}{a_{1}(r^{2})$$

$$-\frac{1}{32}=r^{5}$$

$$r=-\frac{1}{2}$$

Considering the 8th term as the first term, the 14th term corresponds to the 7th term. Then:

- First term: $latex a_{1}=-8$
- Common ratio: $latex r=-\frac{1}{2}$
- Position of term: $latex n=7$

Using the formula, we have:

$latex a_{n}=a_{1}(r^{n-1})$

$$a_{7}=-8(-\frac{1}{2}^{7-1})$$

$$a_{7}=-8(-\frac{1}{2}^6)$$

$$a_{7}=-8(\frac{1}{64})$$

$$a_{7}=-\frac{1}{8}$$

Then, the 14th term of the given sequence is $latex -\frac{1}{8}$.

## Geometric sequences – Practice problems

#### In a geometric sequence, the 4th term is 135 and 7th term is 3645. What is the value of the 15th term?

Write the answer in the input box.

## See also

Interested in learning more about sequences? Take a look at these pages:

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