Geometric Sequences: Formula & Examples

A geometric sequence is a list of numbers where you multiply by the same number each time to get the next term. That number is the common ratio, $r$ . For example, 2, 6, 18, 54, … multiplies by 3. The formula $a_n = a_1 \cdot r^{n-1}$ finds any term.

Before this lesson: You should know arithmetic sequences, exponential functions, and zero and negative exponents.

Extra practice: The Algebra 1 Unit 6 Practice Pack has printable problems, answer keys and a cheat sheet for this unit.

What Is a Geometric Sequence?

A rumor starts with 2 people. Each person tells 3 new people the next day. So 2 people know on day 1, then 6, then 18, then 54. Each day’s number is 3 times the day before.

That is a geometric sequence. An arithmetic sequence adds the same number each time. A geometric sequence multiplies by the same number each time.

To find the common ratio, divide any term by the term right before it. Check that you get the same answer for every pair.

Each term is 3 times the one before it, so the common ratio is r = 3.

Geometric Sequence Formulas

Explicit Formula

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

Here $a_1$ is the first term, $r$ is the common ratio, and $n$ is the term number. To reach the $n$ th term, you multiply by $r$ a total of $n - 1$ times. That is why the exponent is $n - 1$ , not $n$ .

Recursive Formula

a_1 = \text{first term}, \quad a_n = a_{n-1} \cdot r

This says each term is the term before it times $r$ . Like any recursive formula, it must also give the first term.

How to Write the Explicit Formula

Find the first term, $a_1$ .
Find $r$ . Divide each term by the term before it. Check that the answer is the same every time.
Put $a_1$ and $r$ into $a_n = a_1 \cdot r^{n-1}$ .
Do not multiply $a_1$ and $r$ together. The exponent goes on $r$ only.
Check your formula with $n = 1$ and $n = 2$ .

Worked Examples

Example 1: Is It Geometric?

Decide if each sequence is geometric. If it is, find the common ratio.

a) 3, 6, 12, 24, …
b) 2, 5, 8, 11, …
c) 81, −27, 9, −3, …
d) 1, 4, 9, 16, …

\frac{6}{3} = 2, \quad \frac{12}{6} = 2, \quad \frac{24}{12} = 2

Step 1. a) Divide each term by the term before it.

Step 2. The ratios match, so it is geometric with $r = 2$ .

\frac{5}{2} = 2.5, \quad \frac{8}{5} = 1.6

Step 3. b) Divide each term by the term before it.

Step 4. The ratios don’t match, so it is not geometric. It adds 3 each time, so it is arithmetic (see arithmetic sequences).

\frac{-27}{81} = -\tfrac{1}{3}, \quad \frac{9}{-27} = -\tfrac{1}{3}, \quad \frac{-3}{9} = -\tfrac{1}{3}

Step 5. c) Divide each term by the term before it. A negative divided by a positive is negative (see multiplying and dividing integers).

Step 6. The ratios match, so $r = -\tfrac{1}{3}$ . A negative ratio makes the signs switch back and forth.

\frac{4}{1} = 4, \quad \frac{9}{4} = 2.25

Step 7. d) Divide each term by the term before it.

Step 8. The ratios don’t match, so it is not geometric. These are the square numbers.

Answer: a) geometric, $r = 2$ ; b) not geometric; c) geometric, $r = -\tfrac{1}{3}$ ; d) not geometric

Example 2: Write the Formula and Find a Term

Write an explicit formula for 5, 15, 45, … Then find the 8th term.

r = \frac{15}{5} = 3

Step 1. The first term is $a_1 = 5$ . Divide to find the ratio.

a_n = 5 \cdot 3^{n-1}

Step 2. Substitute into $a_n = a_1 \cdot r^{n-1}$ .

Step 3. Check the first term: $a_1 = 5 \cdot 3^0 = 5 \cdot 1 = 5$ . Check the second: $a_2 = 5 \cdot 3^1 = 15$ .

a_8 = 5 \cdot 3^7

Step 4. For the 8th term, substitute $n = 8$ . The exponent is $8 - 1 = 7$ .

a_8 = 5 \cdot 2187

Step 5. Find $3^7 = 2187$ (see exponents).

5 \cdot 2187 = 10{,}935

Step 6. Multiply.

Answer: $a_n = 5 \cdot 3^{n-1}$ , and $a_8 = 10{,}935$ .

Example 3: A Ratio That Is a Fraction

Write a recursive formula and an explicit formula for 64, 32, 16, 8, … Then find the 7th term.

r = \frac{32}{64} = \tfrac{1}{2}

Step 1. Divide to find the ratio.

Step 2. The terms get smaller, because the ratio is between 0 and 1.

a_1 = 64, \quad a_n = a_{n-1} \cdot \tfrac{1}{2}

Step 3. Write the recursive formula: the first term, then the rule for each next term.

a_n = 64\left(\tfrac{1}{2}\right)^{n-1}

Step 4. Write the explicit formula.

a_7 = 64\left(\tfrac{1}{2}\right)^6

Step 5. For the 7th term, substitute $n = 7$ . The exponent is 6.

a_7 = 64 \cdot \tfrac{1}{64}

Step 6. Find the power: $\left(\tfrac{1}{2}\right)^6 = \tfrac{1}{64}$ .

64 \cdot \tfrac{1}{64} = 1

Step 7. Multiply (see multiplying and dividing fractions).

Answer: $a_7 = 1$

Example 4: Find the Term Number

The sequence 2, 6, 18, … keeps going. Which term is 1,458?

a_n = 2 \cdot 3^{n-1}

Step 1. Find the first term and the ratio: $a_1 = 2$ and $r = \frac{6}{2} = 3$ .

2 \cdot 3^{n-1} = 1458

Step 2. Set the formula equal to 1,458.

3^{n-1} = 729

Step 3. Divide both sides by 2.

3^6 = 729

Step 4. Find the power of 3 that makes 729. Multiply 3s until you reach it: 3, 9, 27, 81, 243, 729.

n - 1 = 6

Step 5. The exponents must match.

n = 7

Step 6. Add 1 to both sides.

Answer: 1,458 is the 7th term.

Check: $2 \cdot 3^6 = 2 \cdot 729 = 1458$ .

Example 5: Word Problem

A ball is dropped from 81 feet. Each bounce reaches 2/3 of the height of the bounce before. The first bounce goes up 54 feet. How high is the 4th bounce?

81 \cdot \tfrac{2}{3} = 54

Step 1. The bounce heights form a geometric sequence with $a_1 = 54$ and $r = \tfrac{2}{3}$ . Check the first bounce.

a_n = 54\left(\tfrac{2}{3}\right)^{n-1}

Step 2. Write the explicit formula.

a_4 = 54\left(\tfrac{2}{3}\right)^3

Step 3. For the 4th bounce, substitute $n = 4$ . The exponent is 3.

a_4 = 54 \cdot \tfrac{8}{27}

Step 4. Find the power: $\left(\tfrac{2}{3}\right)^3 = \tfrac{8}{27}$ .

54 \cdot \tfrac{8}{27} = 16

Step 5. Multiply: $\frac{54}{27} = 2$ , and $2 \cdot 8 = 16$ .

Answer: The 4th bounce is 16 feet high. The bounces are 54, 36, 24 and 16 feet.

Common Mistakes

Mistake	Example of the error	Fix
Using $n$ instead of $n - 1$	$a_8 = 5 \cdot 3^8$	$a_8 = 5 \cdot 3^7$
Multiplying $a_1$ and $r$ first	$5 \cdot 3^{n-1} = 15^{n-1}$	The exponent is on 3 only. Do the power first
Subtracting instead of dividing	3, 6, 12: $6 - 3 = 3$ , so $r = 3$	Divide: $\frac{6}{3} = 2$ , so $r = 2$
Dividing in the wrong order	64, 32: $\frac{64}{32} = 2$	$\frac{\text{later term}}{\text{earlier term}}$ : $\frac{32}{64} = \tfrac{1}{2}$
Losing the negative sign	81, −27: $r = \tfrac{1}{3}$	$\frac{-27}{81} = -\tfrac{1}{3}$ . The signs alternate

Geometric Sequences and Exponential Functions

A geometric sequence is an exponential function whose inputs are the term numbers 1, 2, 3, … For example, $a_n = 2 \cdot 3^{n-1}$ gives the same outputs as $f(n) = 2 \cdot 3^{n-1}$ for whole numbers. The graph is a set of separate points on an exponential curve.

	Arithmetic sequence	Geometric sequence
Next term	Add $d$	Multiply by $r$
Example	3, 7, 11, 15, …	3, 6, 12, 24, …
Explicit formula	$a_n = a_1 + (n - 1)d$	$a_n = a_1 \cdot r^{n-1}$
Recursive formula	$a_n = a_{n-1} + d$	$a_n = a_{n-1} \cdot r$
Same kind of function as	Linear	Exponential

Many growth and decay problems are geometric sequences. A balance that earns compound interest once a year makes a geometric sequence with $r = 1 + \text{rate}$ . Adding up the terms of a sequence is a series, which comes in Algebra 2.

Practice Problems

Is 4, 12, 36, 108, … geometric? If so, find r.
Is 10, 5, 5/2, 5/4, … geometric? If so, find r.
Find the next two terms of 2, −8, 32, …
Write an explicit formula for 7, 14, 28, 56, …
Find $a_6$ for $a_n = 3 \cdot 2^{n-1}$ .
Write a recursive formula for 100, 20, 4, …
A geometric sequence has $a_1 = 4$ and $r = -3$ . Find $a_5$ .
A geometric sequence has $a_2 = 12$ and $a_5 = 96$ . Find $r$ and $a_1$ .
Which term of 3, 6, 12, … is 384?
A tournament starts with 64 teams. Half the teams are knocked out in each round. Write a formula for the number of teams in round n. How many teams play in round 6?

Show answers

\frac{12}{4} = 3, \quad \frac{36}{12} = 3, \quad \frac{108}{36} = 3

Divide each term by the one before it.

r = 3

The ratios match. Yes.

5 \div 10 = \tfrac{1}{2}, \quad \tfrac{5}{2} \div 5 = \tfrac{1}{2}, \quad \tfrac{5}{4} \div \tfrac{5}{2} = \tfrac{1}{2}

Divide each term by the one before it.

r = \tfrac{1}{2}

The ratios match. Yes.

r = \frac{-8}{2} = -4

Find the ratio.

32(-4) = -128

Multiply the last term by the ratio.

-128(-4) = 512

Multiply again.

The next two terms are −128 and 512.

a_1 = 7, \quad r = \frac{14}{7} = 2

Find the first term and the ratio.

a_n = 7 \cdot 2^{n-1}

Substitute into the formula.

a_6 = 3 \cdot 2^5

Substitute n = 6. The exponent is 5.

a_6 = 3 \cdot 32

Find $2^5 = 32$ .

3 \cdot 32 = 96

Multiply.

r = \frac{20}{100} = \tfrac{1}{5}

Find the ratio.

a_1 = 100, \quad a_n = a_{n-1} \cdot \tfrac{1}{5}

Write the first term and the rule.

a_5 = 4(-3)^4

Substitute n = 5. The exponent is 4.

(-3)^4 = 81

An even power of a negative is positive.

4 \cdot 81 = 324

Multiply.

12 \cdot r^3 = 96

From the 2nd term to the 5th term is 3 steps, so multiply by r three times.

r^3 = 8

Divide both sides by 12.

r = 2

Find the number that cubes to 8.

a_1 = \frac{12}{2} = 6

Go back one step from the 2nd term to find the 1st.

a_n = 3 \cdot 2^{n-1}

The formula is.

3 \cdot 2^{n-1} = 384

Set it equal to 384.

2^{n-1} = 128

Divide both sides by 3.

2^{n-1} = 2^7

Write 128 as a power of 2.

n - 1 = 7

The exponents match.

n = 8

Add 1 to both sides.

384 is the 8th term.

10.

a_n = 64\left(\tfrac{1}{2}\right)^{n-1}

Start with 64 teams and keep half each round.

a_6 = 64\left(\tfrac{1}{2}\right)^5

Substitute n = 6. The exponent is 5.

\left(\tfrac{1}{2}\right)^5 = \tfrac{1}{32}

Find the power.

64 \cdot \tfrac{1}{32} = 2

Multiply.

2 teams play in round 6 (the final).

Want more practice? The Algebra 1 Unit 6 Practice Pack has 49 pages of problems on this and the other lessons in the unit, with step-by-step answers and a cheat sheet.

FAQ

What is the difference between an arithmetic and a geometric sequence?

An arithmetic sequence adds the same number each time. A geometric sequence multiplies by the same number each time. Check the differences for arithmetic and the ratios for geometric.

How do you find the common ratio?

Divide any term by the term right before it. For 5, 15, 45, … the ratio is $\frac{15}{5} = 3$ . Check at least two pairs.

Can the common ratio be negative or a fraction?

Yes. A negative ratio makes the signs switch, as in 81, −27, 9, −3. A ratio between 0 and 1 makes the terms shrink, as in 64, 32, 16. The ratio can’t be 0.

Is a geometric sequence an exponential function?

Yes. The formula $a_n = a_1 \cdot r^{n-1}$ has the variable in the exponent. The only difference is that the inputs are the whole numbers 1, 2, 3, …

Summary

A geometric sequence multiplies by the same number, the common ratio $r$ , each time.
Find $r$ by dividing a term by the term before it.
Explicit formula: $a_n = a_1 \cdot r^{n-1}$ . Recursive formula: $a_n = a_{n-1} \cdot r$ , with $a_1$ given.
The exponent is $n - 1$ , and it goes on $r$ only.
A geometric sequence is an exponential function with inputs 1, 2, 3, …

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