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Here we will learn what an arithmetic sequence is, how to continue an arithmetic sequence, how to find missing terms in an arithmetic sequence and how to generate an arithmetic sequence.

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

An **arithmetic sequence** is an ordered set of numbers that have a **common difference **between each consecutive term.

For example in the arithmetic sequence 3, 9, 15, 21, 27, the common difference is 6.

An arithmetic sequence can be known as an arithmetic progression. The difference between consecutive terms is an arithmetic sequence is always the same.

If we **add **or** subtract** by the **same number **each time to make the sequence, it is an **arithmetic sequence**.

The** term-to-term rule** tells us how we get from one term to the next.

Here are some examples of arithmetic sequences:

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

3 | Add 6 | 3, 9, 15, 21, 27, … |

8 | Subtract 2 | 8, 6, 4, 2, 0, … |

12 | Add 7 | 12, 19, 26, 33, 40, … |

-4 | Subtract 5 | -4, -9, -14, -19, -24, … |

½ | Add ½ | ½, 1, 1½, 2, 2½, … |

Arithmetic sequences are also known as linear sequences. If we represented an arithmetic sequence on a graph it would form a straight line as it goes up (or down) by the same amount each time. Linear means straight.

In order to **continue an arithmetic series**, you should be able to spot, or calculate, the term-to-term rule. This is done by subtracting two consecutive terms to find the common difference.

The common difference for an arithmetic sequence is the same for every consecutive term and can determine whether a sequence is increasing or decreasing.

- Take two consecutive terms from the sequence.
- Subtract the first term from the next term to find the common difference
d . - Add the common difference to the last term in the sequence to find the next term. Repeat for each new term.

Get your free arithmetic sequence worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREEGet your free arithmetic sequence worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREE**Arithmetic sequence** is part of our series of lessons to support revision on **sequences**. You may find it helpful to start with the main sequences lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include:

The arithmetic sequence formula is:

Where,

a_{n} is the nth term (general term)

a_{1} is the first term

n is the term position

d is the common difference

We get the arithmetic sequence formula by looking at the following example:

We can see the common difference (d) is 6 , so d = 6 .

a_{1} is the first term which is 3

a_{2} is the second term which is 9

a_{3} is the third term which is 15 etc.

However we can write this using the common difference of 6 ,

Calculate the next three terms for the sequence

Take two consecutive terms from the sequence.

Here we will take the numbers

Subtract the first term from the next term to find the common difference, *d*.

*d *= 13 − 10 = 3

Add the common difference to the last term in the sequence to find the next term. Repeat for each new term.

The next three terms in the sequence are

Calculate the next three terms for the sequence

Take two consecutive terms from the sequence.

Here we will take the numbers

Subtract the first term from the next term to find the common difference, *d*.

*d* = -21 − (-15) = -21 + 15 = -6

Add the common difference to the last number in the sequence to find the next term. Repeat for each new term.

The next three terms are

Calculate the next three terms for the sequence

Take two consecutive terms from the sequence.

Here we will take the numbers

Subtract the first term from the next term to find the common difference, *d*.

*d* = 0.9 − 0.7 = 0.2

The next three terms are

Calculate the next three terms for the sequence

\[\frac{1}{2}, \frac{3}{4}, 1, \frac{5}{4}, \frac{3}{2}, \ldots\]

Take two consecutive terms from the sequence.

Here we will take the numbers

\[\frac{5}{4} \text { and } \frac{3}{2}\]

Subtract the first term from the next term to find the common difference, *d*.

\[d=\frac{3}{2}-\frac{5}{4}=\frac{6}{4}-\frac{5}{4}=\frac{1}{4}\]

\[\begin{aligned}
\frac{3}{2}+\frac{1}{4}=\frac{6}{4}+\frac{1}{4}&=\frac{7}{4} \\\\
\frac{7}{4}+\frac{1}{4}=\frac{8}{4}&=2 \\\\
2+\frac{1}{4}=2 \frac{1}{4}&=\frac{9}{4}
\end{aligned}\]

The next three terms are

\[\frac{7}{4}, 2, \text { and } \frac{9}{4}\]

1. Write the next three terms of the sequence 0.22, 0.32, 0.42, 0.52, …

0.42, 0.32, 0.22

0.62, 0.72, 0.82

0.52, 0.62, 0.72

0.63, 0.64, 0.65

The common difference, d = 0.32-0.22 = 0.1 .

0.52+0.1=0.62

0.62+0.1=0.72

0.72+0.1=0.82

2. Calculate the next 3 terms of the sequence 5, 3, 1, -1, -3, …

-5, -7, -9

-5, -3, -1

5, 7, 9

-1, 1, 3

The common difference, d = 3-5 = -2 .

-3+(-2)=-5

-5+(-2)=-7

-7+(-2)=-9

3. By finding the common difference, state the next 3 terms of the sequence -37, -31, -25, -19, -13, …

-7, -1, 5

-7, 1, 7

7, 13, 19

-19, -25, -31

The common difference, d=-31-(-37) = 6 .

-13+6=-7

-7+6=-1

-1+6=5

4. Find the common difference and hence calculate the next three terms of the sequence \frac{3}{4}, \frac{5}{4}, \frac{7}{4}, \frac{9}{4}, \frac{11}{4}, \ldots

Write your solutions as improper fractions.

\frac{12}{4}, \frac{13}{4}, \frac{14}{4}

\frac{13}{5}, \frac{15}{6}, \frac{17}{7}

\frac{13}{4}, \frac{15}{4}, \frac{17}{4}

\frac{9}{4}, \frac{7}{4}, \frac{5}{4}

The common difference,

d=\frac{5}{4} – \frac{3}{4} = \frac{2}{4}

\begin{array}{l} \frac{11}{4} + \frac{2}{4} =\frac{13}{4}\\\\ \frac{13}{4} + \frac{2}{4} =\frac{15}{4}\\\\ \frac{15}{4} + \frac{2}{4} =\frac{17}{4} \end{array}

In order to **find missing numbers in an arithmetic sequence**, we use the common difference. This can be useful when you are asked to find large terms in the sequence and you have been given a consecutive number to the term you are trying to calculate.

- Calculate the common difference between two consecutive terms.
- Add the common difference to the previous term before the missing value.
- Subtract the common difference to the term after a missing value.

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

Find the common difference between two consecutive terms.

*d* = 8 − 5 = 3

Add the common difference to the previous term before the missing value.

Subtract the common difference from the term after a missing value.

The missing terms are

Note: Here, you could repeat Step 2 by using

Find the missing values in the sequence …,

Find the common difference between two consecutive terms.

*d* = -1.2 − (-1.0) = -1.2 + 1 = -0.2

Add the common difference to the previous term before the missing value.

Subtract the common difference from the term after a missing value.

The missing terms are

Find the missing values in the sequence

Find the distance between the two known terms.

Calculate the common difference.

To get from

This distance has a value of

Add the common difference to the first known term until all terms are calculated.

The missing terms are

Find the missing values in the sequence

\[\ldots, \ldots, \frac{15}{16}, 1 \frac{1}{2}, \ldots\]

Write your answers as fractions in their simplest form.

Find the common difference between two consecutive terms.

\[\begin{aligned}
d &=1 \frac{1}{2}-\frac{15}{16} \\\\
&=\frac{3}{2}-\frac{15}{16} \\\\
&=\frac{24}{16}-\frac{15}{16} \\\\
&=\frac{9}{16}
\end{aligned}\]

Add the common difference to the term before the missing value.

\[\begin{aligned}
&1 \frac{1}{2}+\frac{9}{16} \\\\
&=\frac{24}{16}+\frac{9}{16} \\\\
&=\frac{33}{16}=2 \frac{1}{16}
\end{aligned}\]

Subtract the common difference from the term after a missing value.

\[\begin{aligned}
&\frac{15}{16}-\frac{9}{16} \\\\
&=\frac{6}{16}=\frac{3}{8}
\end{aligned}\]

Repeat this step to find the first term in this sequence.

\[\begin{aligned}
&\frac{3}{8}-\frac{9}{16} \\\\
&=\frac{6}{16}-\frac{9}{16} \\\\
&=\frac{-3}{16}
\end{aligned}\]

The missing terms in the sequence are

\[\frac{-3}{16}, \frac{3}{8}, \text { and } 2 \frac{1}{16}\]

1. Find the missing numbers in the arithmetic sequence 7, 14, …, 28, …

20, 34

35, 42

17, 37

21, 35

The common difference, d=14-7=7 .

14+7=21

28+7=35

2. Find the missing numbers in the sequence

\frac{5}{10}, \frac{9}{10}, \ldots, \ldots, 2 \frac{1}{10}

1 \frac{5}{10}, 1 \frac{9}{10}

1 \frac{13}{10}, 1 \frac{17}{10}

1 \frac{3}{10}, 1 \frac{7}{10}

1 \frac{4}{10}, 1 \frac{8}{10}

The common difference,

d= \frac{9}{10} – \frac{5}{10} = \frac{4}{10}

\begin{aligned} \frac{9}{10} + \frac{4}{10} &= \frac{13}{10}\\\\ &=1 \frac{3}{10}\\\\ \frac{13}{10}+\frac{4}{10}&=\frac{17}{10}\\\\ &=1\frac{7}{10} \end{aligned}

3. Find the missing terms in the sequence 1.9, 1.4, …, …, -0.1

0.9, 0.4

1.09, 1.04

1, 0.6

1.39, 1.34

The common difference, d=1.4-1.9 = -0.5 .

1.4+(-0.5)=0.9

0.9+(-0.5)=0.4

4. Calculate the missing terms in the arithmetic sequence …, …, …, -12, -4

12,4, -4

-36, -28, -20

-30, -24, -18

-20, -28, -36

The common difference, d=-4 – – 12 = 8 .

Working backwards:

3rd term: -12-8=-20

2nd term: -20-8=-28

1st term: -28-8=-36

In order to** generate an arithmetic sequence**, we need to know the ** n ^{th} term**.

The

We can work out any number of terms of an arithmetic sequence by substituting values into the ^{th} term.

The first term is found when

the second term when

the fifth term when

the tenth term when

This is known as the** position-to-term rule** as you can calculate the term, given its position in the sequence.

- Find the first term of the sequence by substituting
n = 1 into then ^{th}term. - Find the second term by substituting
n = 2 into then ^{th}term. - Continue to substitute values for
n until all the required terms of the sequence are calculated.

**Top tip:** After you have calculated the first term in the sequence just keep adding the coefficient

Generate the first

Find the first term in the sequence by substituting *n* = 1 into the *n*^{th} term.

When

(5 × 1) − 7 = -2

Find the second term by substituting *n* = 2 into the *n*^{th} term.

When

(5 × 2) − 7 = 10 − 7 = 3

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

When

(5 × 3) − 7 = 15 − 7 = 8

When

(5 × 4) − 7 = 20 − 7 = 13

When

(5 × 5) − 7 = 25 − 7 = 18

The first 5 terms of the sequence

OR

**Top tip:**

When

The coefficient of

Complete the table for the first

1 | 2 | 3 | 4 | 5 | |

Find the first term in the sequence by substituting *n *= 1 into the n^{th} term.

When

6 − 1 = 5.

1 | 2 | 3 | 4 | 5 | |

5 |

Find the second term by substituting *n *= 2 into the n^{th} term.

When

6 − 2 = 4

1 | 2 | 3 | 4 | 5 | |

5 | 4 |

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

When

6 − 3 = 3

When

6 − 4 = 2

When

6 − 5 = 1

1 | 2 | 3 | 4 | 5 | |

5 | 4 | 3 | 2 | 1 |

OR

**Top tip:**

When

6 – 1 = 5

The coefficient of

Red and blue counters are placed into a sequence shown below.

The red counters have an ^{th}

The blue counters have an ^{th}

State the number of red counters in pattern

Calculate the fourth term in the sequence by substituting *n* = 4 into the n^{th} term 2*n*.

When

2 × 4 = 8

There are

Calculate the tenth term by substituting *n* = 10 into the n^{th} term 2*n*.

When

2 × 10 = 20

There are

Substitute the value for *n* into the n^{th} term of the sequence 3*n* − 3.

When

3n − 3 = (3 × 27) − 3 = 81− 3 = 78

There are

The ^{th} term of a sequence is

Find the first term in the sequence by substituting *n* = 1 into the n^{th} term.

When

(3a + b) × 1 = 3a + b

Find the second term by substituting *n *= 2 into the n^{th} term.

When

(3a + b) × 2 = 6a + 2b

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

When

(3a + b) × 3 = 9a + 3b

When

(3a + b) × 4 = 12a +4 b

When

(3a + b) × 5 = 15a + 5b

The first

1. Generate the first 6 terms of the arithmetic sequence 7n-4 .

-4, 3, 10, 17, 24, 31

7, 3, -1, -5, -9, -13

3, 10, 17, 24, 31, 38

1, 8, 15, 22, 29, 36

\begin{aligned}
7 \times 1 – 4 &= 3\\
7 \times 2 – 4 &= 10\\
7 \times 3 – 4 &= 17\\
7 \times 4 – 4 &= 24\\
7 \times 5 – 4 &= 31\\
7 \times 6 – 4 &= 38
\end{aligned}

2. Complete the table to show the first 5 terms of the sequence 2 − 3n .

\begin{aligned} &\quad n \quad \quad 1 \quad \quad 2 \quad \quad 3 \quad \quad 4 \quad \quad 5\\ &2 − 3n \end{aligned}

\begin{aligned}
&\quad n \quad \quad 1 \quad \quad 2 \quad \quad 3 \quad \quad 4 \quad \quad 5\\
&2 − 3n \;\; -1 \quad -4 \quad -7 \;\; -10 \;\; -13
\end{aligned}

\begin{aligned}
&\quad n \quad \quad 1 \quad \quad 2 \quad \quad 3 \quad \quad 4 \quad \quad 5\\
&2 − 3n \;\; -1 \quad -3 \quad -5 \quad -7 \;\; \; -9
\end{aligned}

\begin{aligned}
&\quad n \quad \quad 1 \quad \quad 2 \quad \quad 3 \quad \quad 4 \quad \quad 5\\
&2 − 3n \quad \; 5 \quad \quad 8 \quad \quad 11 \quad \;\; 14 \quad \;17
\end{aligned}

\begin{aligned}
&\quad n \quad \quad 1 \quad \quad 2 \quad \quad 3 \quad \quad 4 \quad \quad 5\\
&2 − 3n \quad \;\: 1 \quad \quad 4 \quad \quad 7 \quad \;\; 10 \quad \;\; 13
\end{aligned}

\begin{aligned}
2-3 \times 1 &= – 1\\
2-3 \times 2 &= – 4\\
2-3 \times 3 &= – 7\\
2-3 \times 4 &= – 10\\
2-3 \times 5 &= – 13
\end{aligned}

3. Calculate the sum of the 1^{st}, 10^{th}, 100^{th} and 1000^{th} term of the sequence 4n-25 .

-60

4344

4404

4119

1st term: 4 × 1-25=-21

10th term: (4 × 10)-25=15

100th term: (4 × 100)-25=375

1000th term: (4 × 1000)-25=3975

-21+15+375+3975=4344

4. Below are the first 3 terms of a pattern. The number of lines is represented by the sequence 4n+1 and the number of triangles is represented by the sequence 2n . How many lines are there in the term with 12 triangles?

49

41

97

25

Since the number of triangles is 2n and there are 12 triangles,

\begin{aligned} 2n&=12\\ n&=6 \end{aligned}

There are 12 triangles in pattern number 6 .

The number of lines is 4n+1 .

When n=6 ,

(4 \times 6) + 1 = 25 .

1. The nth term of a sequence is 4n + 5 .

State the first 5 terms of the sequence.

**(2 marks)**

Show answer

At least 3 terms

**(1****)**

9, 13, 17, 21, 25

**(1)**

2. Work out the missing values in the following sequence:

17, ….., ….., 32, ….

**(2 marks)**

Show answer

\begin{aligned}
d&=\frac{32-17}{3}\\\\
d&=5
\end{aligned}

**(1)**

22, 27, 37

**(1)**

3. Here are the first four terms of an arithmetic sequence

2, 7, 12, 17

Here are the first five terms of another arithmetic sequence

-4, -1, 2, 5, 8

Find two numbers that are in both number sequences.

**(2 marks)**

Show answer

2

**(1)**

17

**(1)**

**Multiplying the value for a term to get another term in the sequence**

E.g.

Let’s look at the sequence4, 10, 16, 22, 28.

The third term in the sequence is16 .

The thirtieth term does not equal the third term multiplied by10, or160 (as16 × 10 = 160 ). The thirtieth term is equal to178 .

**Arithmetic sequences with negative terms do not always decrease**

E.g.

The sequence-48, -40, -32, -24, -16 has a common difference of+8.

This means that even though the sequence is showing negative integers rather than positive integers, it is increasing.

**Adding the constant in the**n ^{th}term instead of the common difference

E.g.

Then ^{th}term3n − 7 will produce a sequence of numbers that have a common difference of3. The misconception would occur if the next term is found by subtracting7 rather than adding3 .

**Simplifying the**n ^{th}term incorrectly

E.g.

Incorrectly simplifying6n + 2 to give8n .

This is incorrect for any value other than whenn = 1 .

You have now learned how to:

- Generate terms of a sequence from either a term-to-term or a position-to-term rule

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