One to one maths interventions built for KS4 success

Weekly online one to one GCSE maths revision lessons now available

This topic is relevant for:

Here we will learn what a **geometric sequence **is, how to continue a geometric sequence, how to find missing terms in a geometric sequence, and how to generate a geometric sequence.

At the end, you’ll find geometric sequence worksheets based on Edexcel, AQA, and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

A geometric sequence (geometric progression) is an ordered set of numbers that progresses by multiplying or dividing each term by a common ratio.

If we **multiply **or** divide **by the **same **number each time to make the sequence, it is a geometric** sequence**.

The common ratio is the same for any two consecutive terms in the same sequence.

Here are a few examples,

First Term | Term-to-Term Rule | First 5 Terms |
---|---|---|

3 | Multiply by 3 | 3, 9, 27, 81, 243, ... |

5 | Multiply by 2 | 5, 10, 20, 40, 80, ... |

-12 | Divide by 2 | -12, -6, -3, -1.5, -0.75, ... |

0.8 | Multiply by 5 | 0.8, 4, 20, 100, 500, ... |

\frac{1}{2} | Divide by 4 | \frac{1}{2}, \; \frac{1}{8}, \; \frac{1}{32}, \; \frac{1}{128}, \; \frac{1}{512} \ldots |

The geometric sequence formula is,

Where,

\pmb{ a_{n} } is the n^{th} term (general term),

\pmb{ a_{1} } is the first term,

\pmb{ n } is the term position,

and \pmb{ r } is the common ratio.

We get the geometric sequence formula by looking at the following example,

We can see the common ratio (r) is 2 , so r = 2 .

a_{1} is the first term which is 5 ,

a_{2} is the second term which is 10 ,

and a_{3} is the third term which is 20 etc.

However we can write this using the common difference of 2 ,

To find out more about the different types of sequences, and how to answer sequence related questions you may find it helpful to look at the introduction to sequences lesson or one of the others in this section.

To continue a geometric sequence, you need to calculate the common ratio. This is the factor that is used to multiply one term to get the next term. To calculate the common ratio and continue a geometric sequence you need to:

**Take two consecutive terms from the sequence.****Divide the second term by the first term to find the common ratio r.****Multiply the last term in the sequence by the common ratio to find the next term. Repeat for each new term.**

Get your free geometric sequences worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREEGet your free geometric sequences worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREECalculate the next three terms for the geometric progression 1, 2, 4, 8, 16, …

**Take two consecutive terms from the sequence.**

Here we will take the numbers 4 and 8 .

2 **Divide the second term by the first term to find the value of the common ratio, r .**

3 **Multiply the last term in the sequence by the common ratio to find the next term. Repeat for each new term.**

The next three terms in the sequence are 32, 64, and 128 .

Calculate the next three terms for the sequence -2, -10, -50, -250, -1250, …

**Take two consecutive terms from the sequence.**

Here we will take the numbers -10 and -50 .

**Divide the second term by the first term to find the common ratio, r. **

r=-50 \div -10
r=5

-1250 \times 5=-6250
-6250 \times 5=-31250
-31250 \times 5=-156250

The next three terms are -6250, -31250, and -156250.

Calculate the next three terms for the sequence 100, 10, 1, 0.1, 0.01, …

**Take two consecutive terms from the sequence.**

Here we will take the numbers 0.1 and 0.01 .

**Divide the second term by the first term to find the common ratio, r.**

\begin{aligned} r&=0.01 \div 0.1 \\\\ r&=0.1 \end{aligned}

\begin{aligned}
0.01 \times 0.1&=0.001\\\\
0.001 \times 0.1&=0.0001\\\\
0.0001 \times 0.1&=0.00001\\\\
\end{aligned}

The next three terms in the sequence are 0.001, 0.0001, and 0.00001.

Calculate the next three terms for the sequence

10, \quad 5, \quad 2\frac{1}{2}, \quad 1 \frac{1}{4}, \quad \frac{5}{8}, ...**Take two consecutive terms from the sequence.**

Here we will take the numbers 5 and 2\frac{1}{2} .

**Divide the second term by the first term to find the common ratio, r. **

\begin{aligned}
r&=2 \frac{1}{2} \div 5\\\\
r&= \frac {1}{2}
\end{aligned}

\begin{aligned}
\frac{5}{8} \times \frac{1}{2}&=\frac{5}{16} \\\\
\frac{5}{16} \times \frac{1}{2}&=\frac{5}{32} \\\\
\frac{5}{32} \times \frac{1}{2}&=\frac{5}{64}
\end{aligned}

The next three terms are

\frac{5}{16}, \frac{5}{32}, and \frac{5}{64} .

1. Write the next three terms of the sequence 0.5, 5, 50, 500, …

1000, 5000, 10000

950, 1400, 1850

5000, 50000, 500000

550, 555, 560

Choose two consecutive terms. For example, 5 and 50 .

Common ratio,

50\div 5=10

500 \times 10=5000, 5000 \times 10=50000, 50000 \times 10=500000

2. Write the next three terms of the sequence 0.04, 0.2, 1, 5, 25, …

50, 100, 200

125, 625, 3125

50, 1000, 20000

12.5, 6.25, 3.125

Choose two consecutive terms. For example, 5 and 25 .

Common ratio,

25\div 5=5

25 \times 5=125, 125 \times 5=625, 625 \times 5=3125

3. Calculate the next 3 terms of the sequence -1, -3, -9, -27, -81, …

243, 729, 2187

-243, -729, -2187

243, -729, 2187

-27, -9, -3

Choose two consecutive terms. For example, -27 and -9 .

Common ratio,

-27\div -9=3

-81 \times 3=-243, -243 \times 3=-729, -729 \times 3=-2187

4. By finding the common ratio, state the next 3 terms of the sequence 640, 160, 40, 10, 2.5 .

0.625, 0.15625, 0.0390625

1.25, 0.625, 0.3125

10, 40, 160

0.25, 0.025, 0.0025

Choose two consecutive terms. For example, 40 and 10 .

Common ratio,

10\div 40=\frac{1}{4} \text{ or } 0.25

2.5 \times 0.25=0.625, 0.625 \times 0.25=0.15625, 0.15625 \times 0.25=0.0390625

5. Work out the common ratio and therefore the next three terms in the sequence 36, 12, 4, \frac{4}{3}, \frac{4}{9}, …

\frac{12}{9}, \frac{36}{9}, \frac{108}{9}

\frac{12}{27}, \frac{36}{81}, \frac{108}{243}

\frac{4}{12}, \frac{4}{15}, \frac{4}{18}

\frac{4}{27}, \frac{4}{81}, \frac{4}{243}

Choose two consecutive terms. For example, 12 and 4 .

Common ratio,

4\div 12=\frac{1}{3}

\frac{4}{9} \times \frac{1}{3}=\frac{4}{27}, \frac{4}{27} \times \frac{1}{3}=\frac{4}{81}, \frac{4}{81} \times \frac{1}{3}=\frac{4}{243}

6. Find the common ratio and hence calculate the next three terms of the sequence 1, -1, 1, -1, 1, …

1, 1, 1

-1, -1, -1

1, -1, 1

-1, 1, -1

Choose two consecutive terms. For example, -1 and 1 .

Common ratio,

-1\div 1=-1

1 \times -1=-1, -1 \times -1=1, 1 \times -1=-1

The common ratio can be used to find missing numbers in a geometric sequence. To find missing numbers in a geometric sequence you need to:

**Calculate the common ratio between two consecutive terms.****Multiply the term before any missing value by the common ratio.****Divide the term after any missing value by the common ratio.**

Repeat Steps 2 and 3 until all missing values are calculated. You may only need to use Step 2 or 3 depending on what terms you have been given.

Fill in the missing terms in the sequence 7, 14, …, …,112 .

**Calculate the common ratio between two consecutive terms.**

\begin{aligned}
r&=14 \div 7 \\\\
r&=2
\end{aligned}

**Multiply the term before any missing value by the common ratio.**

14 \times 2=28

**Divide the term after any missing value by the common ratio.**

112 \div 2 = 56

The missing terms are 28 and 56 .

**Note:** Here, you could repeat Step 2 by using 28 \times 2 = 56.

Find the missing values in the sequence 0.4, …, ..., 137.2, 960.4.

**Calculate the common ratio between two consecutive terms.**

\begin{aligned}
r&=960.4 \div 137.2 \\\\
r&=7
\end{aligned}

**Multiply the term before any missing value by the common ratio.**

0.4 \times 7 = 2.8

**Divide the term after any missing value by the common ratio.**

137.2 \div 7 = 19.6

The missing terms are 2.8 and 19.6 .

Find the missing values in the sequence, -4, ..., …, -108,...

**Calculate the common ratio between two consecutive terms.**

First, we need to find the factor between the two terms, -108 \div -4 = 27 .

To get from -108 to -4 , we jump 3 terms.

This means that -4 has been multiplied by the common ratio three times or -4 \times r \times r \times r = -4r^3 .

\begin{aligned} r^{3}&=27\\\\ r&=3 \end{aligned}

**Multiply the term before any missing value by the common ratio.**

\begin{aligned} -4 \times 3&=-12 \\\\ -12 \times 3&=-36 \\\\ -36 \times 3&=-108 \\\\ -108 \times 3&=-324 \end{aligned}

**Note:** Term -108 is already given.

The missing terms are -12, -36, and -324.

**Divide the term after any missing value by the common ratio.**

We don’t need to complete this step.

Find the missing values in the sequence

…, 4 \frac{1}{2}, \quad 13 \frac{1}{2},\quad ... , \quad... **Calculate the common ratio between two consecutive terms.**

\begin{aligned}
r&=13 \frac{1}{2} \div 4 \frac{1}{2}\\\\
r &=\frac{27}{2} \div \frac{9}{2}\\\\
r &=\frac{27}{2} \times \frac{2}{9}\\\\
r &=3
\end{aligned}

**Multiply the term before any missing value by the common ratio.**

13 \frac{1}{2} \times 3=40 \frac{1}{2}

Repeat this step to find the next term.

40 \frac{1}{2} \times 3=121 \frac{1}{2}

**Divide the term after any missing value by the common ratio.**

4 \frac{1}{2} \div 3=1 \frac{1}{2}

The missing terms in the sequence are

1 \frac{1}{2}, 40 \frac{1}{2}, and 121 \frac{1}{2} .

1. Find the missing numbers in the geometric sequence 4, 2, …, 0.5, …

1, 0.2

1, 0.25

1, 0.1

1, 0

Choose two consecutive terms. For example, 4 and 2 .

Common ratio,

2\div 4=0.5

2 \times 0.5=1, 0.5 \times 0.5=0.25

2. Find the missing numbers in the sequence -7, -35, …, …, -4375

-175, -875

175, 875

-4375, -546875

1481.7, 2928.3

Choose two consecutive terms. For example, -7 and -35 .

Common ratio,

-35\div -7=5

-35 \times 5=-175, -175 \times 5=-875

3. Find the missing terms in the sequence 0.6, …, …, 0.075, 0.0375

0.2, 0.067

1.2, 2.4

0.12, 0.24

0.3, 0.15

Choose two consecutive terms. For example, 0.075 and 0.0375 .

Common ratio,

0.0375\div 0.075=0.5

-35 \times 5=-175, -175 \times 5=-875

4. Calculate the missing terms in the arithmetic sequence 2 \frac{1}{5}, \frac{11}{20}, \frac{11}{80}, \ldots, \ldots

\frac{11}{320}, \frac{11}{1280}

\frac{44}{80}, \frac{176}{80}

\frac{44}{320}, \frac{176}{1280}

\frac{11}{140}, \frac{11}{200}

Choose two consecutive terms. For example, \frac{11}{20} and \frac{11}{80} .

Common ratio,

\frac{11}{80} \div \frac{11}{20}=\frac{11}{80} \times \frac{20}{11}=\frac{1}{4}

\frac{11}{80} \times \frac{1}{4}=\frac{11}{320}, \frac{11}{320} \times \frac{1}{4}=\frac{11}{1280}

5. Work out the missing terms in the sequence 3, …, …, 24 .

10, 17

9, 18

6, 12

12, 18

3 has been multiplied by the common ratio, r, three times to get 24.

Therefore,

3 \times r \times r \times r=24 \text{ or } 3r^{3}=24 .

Solving the equation,

\begin{aligned} 3r^{3}&=24\\ r^{3}&=8\\ r&=2 \end{aligned}

3 \times 2=6, 6 \times 2=12

6. Work out the missing terms in the sequence 90, …, …, \frac{10}{3} .

\frac{550}{9}, \frac{290}{9}

9, 3

30, 10

45, \frac{45}{2}

90 has been multiplied by the common ratio, r, three times to get \frac{10}{3}.

Therefore,

90 \times r \times r \times r=\frac{10}{3} \text{ or } 90r^{3}=\frac{10}{3} .

Solving the equation,

\begin{aligned} 90r^{3}&=\frac{10}{3}\\ r^{3}&=\frac{10}{3} \div 90\\ r^{3} = \frac{1}{27} r&=\frac{1}{3} \end{aligned}

90 \times \frac{1}{3}=30, 30 \times \frac{1}{3}=10

In order to generate a geometric sequence, we need to know the n^{th} term. Using a as the first term of the sequence, r as the common ratio and n to represent the position of the term, the n^{th} term of a geometric sequence is written as ar^{n-1}.

Once we know the first term and the common ratio, we can work out any number of terms in the sequence.

The first term is found when n=1 , the second term when n=2 , the third term when n=3 and so on.

To generate a geometric sequence you need to:

**Substitute n=1 into the n^{th} term to calculate the first term.****Substitute n=2 into the n^{th} term to calculate the second term.****Continue to substitute values for n until all the required terms of the sequence are calculated.**

Generate the first 5 terms of the sequence 4^{n-1} .

**Substitute n = 1 into the n^{th} term to calculate the first term.**

When n = 1,\quad 4^{1-1} = 4^{0} = 1 .

**Substitute n = 2 into the n^{th} term to calculate the second term.**

When n = 2,\quad 4^{2-1 }= 4^{1} = 4 .

**Continue to substitute values for n until all the required terms of the sequence are calculated.**

When n=3, \quad 4^{3-1}=4^{2}=16 .

When n=4, \quad 4^{4-1}=4^{3}=64 .

When n=5, \quad 4^{5-1}=4^{4}=256 .

The first 5 terms of the sequence are 1, 4, 16, 64, 256.

Complete the table for the first 5 terms of the arithmetic sequence 6 \times 2^{n-1}.

n | 1 | 2 | 3 | 4 | 5 |

2^{n-1} | |||||

6 × 2^{n-1} |

**Substitute n = 1 into the n^{th} term to calculate the first term.**

n | 1 | 2 | 3 | 4 | 5 |

2^{n-1} | 2^{0} = 1 | ||||

6 × 2^{n-1} | 6 |

**Substitute n = 2 into the n^{th} term to calculate the second term.**

n | 1 | 2 | 3 | 4 | 5 |

2^{n-1} | 2^{0} = 1 | 2^{1} = 2 | |||

6 × 2^{n-1} | 6 | 12 |

**Continue to substitute values for n until all the required terms of the sequence are calculated.**

n | 1 | 2 | 3 | 4 | 5 |

2^{n-1} | 2^{0} = 1 | 2^{1} = 2 | 2^{2} = 4 | 2^{3} = 8 | 2^{4} = 16 |

6 × 2^{n-1} | 6 | 12 | 24 | 48 | 96 |

A geometric sequence has the n^{th} term \left(\frac{1}{2}\right)^{n} .

Calculate the 1^{st}, 2^{nd}, 10^{th} and 12^{th} terms in the sequence. Express your answers as fractions.

**Substitute n = 1 into the n^{th} term to calculate the first term.**

When n=1,\quad \left(\frac{1}{2}\right)^{1}=\frac{1}{2} .

**Substitute n = 2 into the n^{th} term to calculate the second term.**

When n=2, \quad \left(\frac{1}{2}\right)^{2}=\frac{1}{4} .

**Continue to substitute values for n until all the required terms of the sequence are calculated.**

When n=10,\quad \left(\frac{1}{2}\right)^{10}=\frac{1}{1024} .

When n=12,\quad \left(\frac{1}{2}\right)^{12}=\frac{1}{4096} .

The unknown terms are

1, \frac{1}{4}, \frac{1}{1024}, and \frac{1}{4096} .

Generate the first 5 terms of the geometric sequence 2(- 3)^{n-1} .

**Substitute n = 1 into the n^{th} term to calculate the first term.**

When n=1, \quad 2(-3)^{n-1}=2(-3)^{1-1}=2(-3)^{0}=2 \times 1=2 .

**Substitute n = 2 into the n^{th} term to calculate the second term.**

When n=2,\quad 2(-3)^{n-1}=2(-3)^{2-1}=2(-3)^{1}=2 \times-3=-6 .

**Continue to substitute values for n until all the required terms of the sequence are calculated.**

When n=3, \quad 2(-3)^{n-1}=2(-3)^{3-1}=2(-3)^{2}=2 \times 9=18 .

When n=4, \quad 2(-3)^{n-1}=2(-3)^{4-1}=2(-3)^{3}=2 \times-27=-54 .

When n=5, \quad 2(-3)^{n-1}=2(-3)^{5-1}=2(-3)^{4}=2 \times 81=162 .

The first 5 terms of the sequence are 2, -6, 18, -54, and 162 .

1. Generate the first 5 terms of the sequence 10^{n} .

10, 100, 1000, 10000, 100000

1, 10, 100, 1000, 10000

10, 20, 30, 40, 50

100, 1000, 10000, 100000, 1000000

When n=1, 10^{1}=10 .

When n=2, 10^{2}=100 .

When n=3, 10^{3}=1000 .

When n=4, 10^{4}=10000 .

When n=5, 10^{5}=100000 .

2. Generate the first 5 terms of the sequence 5^{n-1} .

5, 25, 125, 625, 3125

0.2, 1, 5, 25, 125

625, 125, 25, 5, 1

1, 5, 25, 125, 625

When n=1, 5^{1-1}=5^{0}=1 .

When n=2, 5^{2-1}=5^{1}=5 .

When n=3, 5^{3-1}=5^{2}=25 .

When n=4, 5^{4-1}=5^{3}=125 .

When n=5, 5^{5-1}=5^{4}=625 .

3. Generate the first 5 terms of the sequence 4 \times 3^{n-1} .

1, 3, 9, 27, 81

1, 12, 144, 1728, 20736

4, 12, 36, 108, 324

12, 36, 108, 324, 972

When n=1, 4 \times 3^{1-1}=4 \times 3^{0}=4 .

When n=2, 4 \times 3^{2-1}=4 \times 3^{1}=12 .

When n=3, 4 \times 3^{3-1}=4 \times 3^{2}=36 .

When n=4, 4 \times 3^{4-1}=4 \times 3^{3}=108 .

When n=5, 4 \times 3^{5-1}=4 \times 3^{4}=324 .

4. Generate the first 5 terms of the sequence \frac{3^{n}}{6} .

\frac{1}{2}, 1 \frac{1}{2}, 4 \frac{1}{2}, 13 \frac{1}{2}, 40 \frac{1}{2}

\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \frac{1}{32}

\frac{1}{6}, \frac{1}{2}, 1 \frac{1}{2}, 4 \frac{1}{2}, 13 \frac{1}{2}

\frac{1}{6}, 1 \frac{1}{2}, 4 \frac{1}{2}, 13 \frac{1}{2}, 40 \frac{1}{2}

When n=1, \frac{3^1}{6}= \frac{1}{2} .

When n=2, \frac{3^2}{6}= \frac{9}{6} = 1 \frac{1}{2} .

When n=3, \frac{3^3}{6}= \frac{27}{6} = 4 \frac{1}{2} .

When n=4, \frac{3^4}{6}= \frac{81}{6} = 13 \frac{1}{2} .

When n=5, \frac{3^5}{6}= \frac{243}{6} = 40 \frac{1}{2} .

5. Calculate the 1st, 3rd, 10th and 15th term of the sequence 2^{n}.

2, 4, 8, 16

2, 8, 1024, 32768

1, 4, 512, 16384

4, 16, 2048, 65536

When n=1, 2^{1}= 2 .

When n=3, 2^{3}= 8 .

When n=10, 2^{10}= 1024 .

When n=15, 2^{15}= 32768 .

6. Calculate the first 5 terms of the sequence 3 \times (-5)^{n-1}.

3, -15, 75, -375, 1875

1, -5, 25, -125, 625

1, -225, 3375, -50625, 759375

-15, 75, -375, 1875, -9375

When n=1, 3 \times (-5)^{1-1} = 3 \times (-5)^{0} = 3 .

When n=2, 3 \times (-5)^{2-1} = 3 \times (-5)^{1} = -15 .

When n=3, 3 \times (-5)^{3-1} = 3 \times (-5)^{2} = 75 .

When n=4, 3 \times (-5)^{4-1} = 3 \times (-5)^{3} = -375 .

When n=5, 3 \times (-5)^{5-1} = 3 \times (-5)^{4} = 1875.

1. Which sequence is a geometric progression?

1, 3, 5, 7, 9,…. \quad \quad \quad 1, 3, 9, 27, 81, …..

1, 3, 6, 10, 15, …. \quad \quad 1, 0.6, 0.2, -0.2, -0.6,….

**(1 mark)**

Show answer

1, 3, 9, 27, 81, …..

**(1)**

2. Here is a geometric progression,

1, -5, 25, …., 625, …

(a) Find the common ratio.

(b) Work out the fourth term of the sequence.

**(3 marks)**

Show answer

(a)

25 \div -5 = – 5

Common ratio = -5

**(1)**

(b)

25 \times -5

**(1)**

= -125

**(1)**

3. A scientist is studying a type of bacteria. The number of bacteria over the first four days is shown below.

Day 1 | Day 2 | Day 3 | Day 4 |

60 | 180 | 540 | 1620 |

How many bacteria will there be on day 7?

**(3 marks)**

Show answer

180 \div 60 = 3

**(1)**

1620 \times 3 \times 3 \times 3

**(1)**

43740

**(1)**

**Mixing up the common ratio with the common difference for arithmetic sequences**

Although these two phrases are similar, each successive term in a geometric sequence of numbers is calculated by multiplying the previous term by a common ratio and not by adding a common difference.

**A negative value for**r means that all terms in the sequence are negative

This is not always the case as when r is raised to an even power, the solution is always positive.

**The first term in a geometric sequence**

The first term is a . With ar^{n-1} , the first term would occur when n = 1 and so the power of r would be equal to 0 . Anything to the power of 0 is equal to 1 , leaving a as the first term in the sequence. This is usually mistaken when a = 1 as it is not clearly noted in the question for example, 2^{n-1} is the same as 1 \times 2^{n-1} .

**Incorrect simplifying of the n**^{th}term

For example, 6 \times 3^{n-1} is incorrectly simplified to 18^{n-1} as 6 \times 3 = 18 .

**The difference between an arithmetic and a geometric sequence**

Arithmetic sequences are formed by adding or subtracting the same number. Geometric sequences are formed by multiplying or dividing the same number.

You have now learned how to:

- Recognise geometric sequences

Prepare your KS4 students for maths GCSEs success with Third Space Learning. Weekly online one to one GCSE maths revision lessons delivered by expert maths tutors.

Find out more about our GCSE maths revision programme.