1.4.8 Arithmetic sequences and series (A-level only)

What you'll learn

How to recognise an arithmetic sequence and find its common difference.
How to find any term using the formula for the nth term.
How to calculate the sum of the first $n$ terms efficiently.
How to form and solve equations involving arithmetic sequences and series.

Sequences and series

A sequence is an ordered list of numbers called terms. The position of each term matters.

For example,

13,…4,\ 7,\ 10,\ 13,\ldots

is a sequence. Its first term is 4, its second term is 7, and so on.

We write $u_n$ for the term in position $n$ . Thus, $u_1$ is the first term, $u_2$ is the second term and $u_{20}$ is the twentieth term.

A series is produced by adding the terms of a sequence. The corresponding series is

4+7+10+13+⋯4+7+10+13+\cdots

Definition

Sequence and series

A sequence is an ordered list of terms. A series is the sum of terms from a sequence.

Arithmetic sequences

An arithmetic sequence is a sequence in which the difference between each term and the previous term is constant. This constant is called the common difference, written as $d$ .

For example, in

32,…11,\ 18,\ 25,\ 32,\ldots

the common difference is $d = 7$ because 7 is added each time.

The common difference can be negative. For example,

8,…20,\ 16,\ 12,\ 8,\ldots

has $d = - 4$ .

Key Idea

Recognising an arithmetic sequence

Subtract consecutive terms in the same order. A sequence is arithmetic if every calculation of $u_{n+1}-u_n$ gives the same value.

Example

Checking whether a sequence is arithmetic

Determine whether $23,…5,\ 11,\ 17,\ 23,\ldots$ is arithmetic and state its common difference.

Subtract each term from the term immediately after it:
$11−5=6,17−11=6,23−17=611-5=6,\qquad 17-11=6,\qquad 23-17=6$
The differences are all equal, so the sequence is arithmetic.
The common difference is therefore $d = 6$ .

Common Mistake

Reversing the subtraction

Calculate later term minus earlier term. For a decreasing sequence, the common difference must be negative.

Finding the nth term

Suppose an arithmetic sequence has first term $a$ and common difference $d$ .

Starting from $a$ :

the second term is $a + d$ ;
the third term is $a + 2 d$ ;
the fourth term is $a + 3 d$ .

To reach term $n$ from the first term, you make $n - 1$ equal jumps. Therefore:

u_n=a+(n-1)d

Definition

The nth-term formula

For an arithmetic sequence with first term $a$ and common difference $d$ , the term in position $n$ is

u_n=a+(n-1)d.

Example

Finding a specified term

Find the twenty-fifth term of the arithmetic sequence

23,…8,\ 13,\ 18,\ 23,\ldots

Identify the first term and common difference:
$a=8,d=5a=8,\qquad d=5$
Substitute $n = 25$ into the nth-term formula:
$u_{25}=8+(25-1)(5)$
Calculate the term:
$u_{25}=8+120=128$

Common Mistake

Using n differences

The nth term is reached after $n - 1$ jumps from the first term, so the formula contains $(n - 1) d$ , not $n d$ .

Writing the nth term in a simpler form

You will often simplify $a + (n - 1) d$ into a linear expression in $n$ .

For example, if $a = 12$ and $d = - 3$ , then

un=12+(n−1)(−3)=12−3n+3=15−3n.\begin{aligned} u_n&=12+(n-1)(-3)\\ &=12-3n+3\\ &=15-3n. \end{aligned}

This form makes it easier to solve equations involving particular terms.

Finding the position of a term

If you know a term's value but not its position, set the nth-term formula equal to that value and solve for $n$ .

Example

Finding where a term occurs

The sequence $22,…7,\ 12,\ 17,\ 22,\ldots$ is arithmetic. Determine whether 157 is a term, and find its position if it is.

Identify $a = 7$ and $d = 5$ , then form the nth term:
$u_n=7+(n-1)5=5n+2$
Set the nth term equal to 157:
$5 n + 2 = 157$
Solve the equation:
$5n=155⇒n=315n=155\Rightarrow n=31$
Since 31 is a positive integer, 157 is the thirty-first term.

Common Mistake

A term position must be valid

A solution for $n$ represents a term position only if it is a positive integer. For example, $n = 6.4$ would show that the proposed value is not a term of the sequence.

Using information about different terms

Sometimes the first term and common difference are unknown, but two other terms are given. Write each condition using $u_n=a+(n-1)d$ , then solve the simultaneous equations.

Example

Finding a sequence from two terms

An arithmetic sequence has $u_5=18$ and $u_{12}=46$ . Find $a$ and $d$ .

Translate each given term into an equation:
$a+4d=18,a+11d=46a+4d=18,\qquad a+11d=46$
Subtract the first equation from the second to eliminate $a$ :
$7d=28⇒d=47d=28\Rightarrow d=4$
Substitute $d = 4$ into $a + 4 d = 18$ :
$a+16=18⇒a=2a+16=18\Rightarrow a=2$
The sequence therefore has first term 2 and common difference 4, with nth term $u_n=4n-2$ .

Tip

Differences between distant terms

From term $u_p$ to term $u_q$ , there are $q - p$ jumps. Therefore $u_q-u_p=(q-p)d$ , which can give the common difference quickly.

Arithmetic series

The sum of the first $n$ terms of a sequence is written as $S_n$ .

For an arithmetic sequence,

Sn=a+(a+d)+(a+2d)+⋯+[a+(n−1)d].S_n=a+(a+d)+(a+2d)+\cdots+\left[a+(n-1)d\right].

The last term is

l = a + (n - 1) d,

where $l$ denotes the last term being included.

Deriving the sum formula

Write the sum once forwards and once backwards:

Sn=a+(a+d)+⋯+(l−d)+lSn=l+(l−d)+⋯+(a+d)+a.\begin{aligned} S_n&=a+(a+d)+\cdots+(l-d)+l\\ S_n&=l+(l-d)+\cdots+(a+d)+a. \end{aligned}

Adding vertically gives $n$ identical pairs, each equal to $a + l$ :

2S_n=n(a+l).

Therefore,

Sn=n2(a+l).S_n=\frac{n}{2}(a+l).

Since $l = a + (n - 1) d$ , this can also be written as

Sn=n2[2a+(n−1)d].S_n=\frac{n}{2}\left[2a+(n-1)d\right].

Key Idea

Two forms of the sum formula

Use

Sn=n2(a+l)S_n=\frac{n}{2}(a+l)

when the first and last terms are known, or

Sn=n2[2a+(n−1)d]S_n=\frac{n}{2}\left[2a+(n-1)d\right]

when the first term and common difference are known.

Example

Summing the first terms of a sequence

Find the sum of the first 40 terms of the arithmetic sequence $13,…3,\ 8,\ 13,\ldots$

Identify $a = 3$ , $d = 5$ and $n = 40$ .
Use the form involving the common difference:
$S40=402[2(3)+(40−1)5]S_{40}=\frac{40}{2}\left[2(3)+(40-1)5\right]$
Simplify:
$S40=20(6+195)=20(201)=4020.\begin{aligned} S_{40}&=20(6+195)\\ &=20(201)\\ &=4020. \end{aligned}$

Common Mistake

Confusing a term with a sum

$u_n$ is one term, whereas $S_n$ is the total of the first $n$ terms. Check which quantity the question asks for before choosing a formula.

Finding the number of terms from a sum

When $S_n$ is given, substituting into the sum formula often produces a quadratic equation. Solve it and then check that $n$ is a positive integer.

Example

Finding how many terms are included

The arithmetic series $5+8+11+⋯5+8+11+\cdots$ has sum 1025. Find the number of terms.

Identify $a = 5$ , $d = 3$ and $S_n=1025$ , then substitute:
$1025=n2[2(5)+(n−1)3]1025=\frac{n}{2}\left[2(5)+(n-1)3\right]$
Simplify to form a quadratic equation:
$2050=n(3n+7)3n2+7n−2050=0.\begin{aligned} 2050&=n(3n+7)\\ 3n^2+7n-2050&=0. \end{aligned}$
Factorise:
$(3 n + 82) (n - 25) = 0$
The solutions are $n=−823n=-\frac{82}{3}$ and $n = 25$ . Reject the negative solution because a number of terms must be a positive integer, so 25 terms are included.

Modelling with arithmetic sequences

Arithmetic sequences can model quantities that change by a fixed amount in equal stages, such as annual payments increasing by the same amount each year.

Before using the formulas, decide what one term represents and match the first stage to $n = 1$ .

Example

Modelling increasing annual payments

A payment is £600 in the first year and rises by £45 each year. Find the payment in year 10 and the total paid over the first 10 years.

Model the yearly payments as an arithmetic sequence with $a = 600$ , $d = 45$ and $n = 10$ .
Find the tenth payment:
$u_{10}=600+(10-1)45=1005$
The payment in year 10 is £1005.
Find the total using the first and last payments:
$S10=102(600+1005)=8025S_{10}=\frac{10}{2}(600+1005)=8025$
The total paid is £8025.

Exam technique

In the exam

Write down $a$ , $d$ and $n$ before substituting, paying particular attention to the sign of $d$ .
Use $u_n=a+(n-1)d$ for an individual term and a formula for $S_n$ only when terms are being added.
Check that any calculated term position or number of terms is a positive integer, and interpret the answer in its context.
Keep exact values throughout unless the question explicitly asks for a decimal approximation.

Self review

Check yourself

How would you find the common difference and nth term of $11,…19,\ 15,\ 11,\ldots$ ?
An arithmetic sequence has $u_4=17$ and $u_{10}=41$ . Can you find its first term and common difference?
Which form of the sum formula would be most efficient if you knew the first term, last term and number of terms?

What you'll learn

How to recognise an arithmetic sequence and find its common difference.
How to find any term using the formula for the nth term.
How to calculate the sum of the first $n$ terms efficiently.
How to form and solve equations involving arithmetic sequences and series.

Sequences and series

A sequence is an ordered list of numbers called terms. The position of each term matters.

For example,

13,…4,\ 7,\ 10,\ 13,\ldots

is a sequence. Its first term is 4, its second term is 7, and so on.

We write $u_n$ for the term in position $n$ . Thus, $u_1$ is the first term, $u_2$ is the second term and $u_{20}$ is the twentieth term.

A series is produced by adding the terms of a sequence. The corresponding series is

4+7+10+13+⋯4+7+10+13+\cdots

Definition

Sequence and series

A sequence is an ordered list of terms. A series is the sum of terms from a sequence.

Arithmetic sequences

An arithmetic sequence is a sequence in which the difference between each term and the previous term is constant. This constant is called the common difference, written as $d$ .

For example, in

32,…11,\ 18,\ 25,\ 32,\ldots

the common difference is $d = 7$ because 7 is added each time.

The common difference can be negative. For example,

8,…20,\ 16,\ 12,\ 8,\ldots

has $d = - 4$ .

Key Idea

Recognising an arithmetic sequence

Subtract consecutive terms in the same order. A sequence is arithmetic if every calculation of $u_{n+1}-u_n$ gives the same value.

Example

Checking whether a sequence is arithmetic

Determine whether $23,…5,\ 11,\ 17,\ 23,\ldots$ is arithmetic and state its common difference.

Subtract each term from the term immediately after it:
$11−5=6,17−11=6,23−17=611-5=6,\qquad 17-11=6,\qquad 23-17=6$
The differences are all equal, so the sequence is arithmetic.
The common difference is therefore $d = 6$ .

Common Mistake

Reversing the subtraction

Calculate later term minus earlier term. For a decreasing sequence, the common difference must be negative.

Finding the nth term

Suppose an arithmetic sequence has first term $a$ and common difference $d$ .

Starting from $a$ :

the second term is $a + d$ ;
the third term is $a + 2 d$ ;
the fourth term is $a + 3 d$ .

To reach term $n$ from the first term, you make $n - 1$ equal jumps. Therefore:

u_n=a+(n-1)d

Definition

The nth-term formula

For an arithmetic sequence with first term $a$ and common difference $d$ , the term in position $n$ is

u_n=a+(n-1)d.

Example

Finding a specified term

Find the twenty-fifth term of the arithmetic sequence

23,…8,\ 13,\ 18,\ 23,\ldots

Identify the first term and common difference:
$a=8,d=5a=8,\qquad d=5$
Substitute $n = 25$ into the nth-term formula:
$u_{25}=8+(25-1)(5)$
Calculate the term:
$u_{25}=8+120=128$

Common Mistake

Using n differences

The nth term is reached after $n - 1$ jumps from the first term, so the formula contains $(n - 1) d$ , not $n d$ .

Writing the nth term in a simpler form

You will often simplify $a + (n - 1) d$ into a linear expression in $n$ .

For example, if $a = 12$ and $d = - 3$ , then

un=12+(n−1)(−3)=12−3n+3=15−3n.\begin{aligned} u_n&=12+(n-1)(-3)\\ &=12-3n+3\\ &=15-3n. \end{aligned}

This form makes it easier to solve equations involving particular terms.

Finding the position of a term

If you know a term's value but not its position, set the nth-term formula equal to that value and solve for $n$ .

Example

Finding where a term occurs

The sequence $22,…7,\ 12,\ 17,\ 22,\ldots$ is arithmetic. Determine whether 157 is a term, and find its position if it is.

Identify $a = 7$ and $d = 5$ , then form the nth term:
$u_n=7+(n-1)5=5n+2$
Set the nth term equal to 157:
$5 n + 2 = 157$
Solve the equation:
$5n=155⇒n=315n=155\Rightarrow n=31$
Since 31 is a positive integer, 157 is the thirty-first term.

Common Mistake

A term position must be valid

A solution for $n$ represents a term position only if it is a positive integer. For example, $n = 6.4$ would show that the proposed value is not a term of the sequence.

Using information about different terms

Sometimes the first term and common difference are unknown, but two other terms are given. Write each condition using $u_n=a+(n-1)d$ , then solve the simultaneous equations.

Example

Finding a sequence from two terms

An arithmetic sequence has $u_5=18$ and $u_{12}=46$ . Find $a$ and $d$ .

Translate each given term into an equation:
$a+4d=18,a+11d=46a+4d=18,\qquad a+11d=46$
Subtract the first equation from the second to eliminate $a$ :
$7d=28⇒d=47d=28\Rightarrow d=4$
Substitute $d = 4$ into $a + 4 d = 18$ :
$a+16=18⇒a=2a+16=18\Rightarrow a=2$
The sequence therefore has first term 2 and common difference 4, with nth term $u_n=4n-2$ .

Tip

Differences between distant terms

From term $u_p$ to term $u_q$ , there are $q - p$ jumps. Therefore $u_q-u_p=(q-p)d$ , which can give the common difference quickly.

Arithmetic series

The sum of the first $n$ terms of a sequence is written as $S_n$ .

For an arithmetic sequence,

Sn=a+(a+d)+(a+2d)+⋯+[a+(n−1)d].S_n=a+(a+d)+(a+2d)+\cdots+\left[a+(n-1)d\right].

The last term is

l = a + (n - 1) d,

where $l$ denotes the last term being included.

Deriving the sum formula

Write the sum once forwards and once backwards:

Sn=a+(a+d)+⋯+(l−d)+lSn=l+(l−d)+⋯+(a+d)+a.\begin{aligned} S_n&=a+(a+d)+\cdots+(l-d)+l\\ S_n&=l+(l-d)+\cdots+(a+d)+a. \end{aligned}

Adding vertically gives $n$ identical pairs, each equal to $a + l$ :

2S_n=n(a+l).

Therefore,

Sn=n2(a+l).S_n=\frac{n}{2}(a+l).

Since $l = a + (n - 1) d$ , this can also be written as

Sn=n2[2a+(n−1)d].S_n=\frac{n}{2}\left[2a+(n-1)d\right].

Key Idea

Two forms of the sum formula

Use

Sn=n2(a+l)S_n=\frac{n}{2}(a+l)

when the first and last terms are known, or

Sn=n2[2a+(n−1)d]S_n=\frac{n}{2}\left[2a+(n-1)d\right]

when the first term and common difference are known.

Example

Summing the first terms of a sequence

Find the sum of the first 40 terms of the arithmetic sequence $13,…3,\ 8,\ 13,\ldots$

Identify $a = 3$ , $d = 5$ and $n = 40$ .
Use the form involving the common difference:
$S40=402[2(3)+(40−1)5]S_{40}=\frac{40}{2}\left[2(3)+(40-1)5\right]$
Simplify:
$S40=20(6+195)=20(201)=4020.\begin{aligned} S_{40}&=20(6+195)\\ &=20(201)\\ &=4020. \end{aligned}$

Common Mistake

Confusing a term with a sum

$u_n$ is one term, whereas $S_n$ is the total of the first $n$ terms. Check which quantity the question asks for before choosing a formula.

Finding the number of terms from a sum

When $S_n$ is given, substituting into the sum formula often produces a quadratic equation. Solve it and then check that $n$ is a positive integer.

Example

Finding how many terms are included

The arithmetic series $5+8+11+⋯5+8+11+\cdots$ has sum 1025. Find the number of terms.

Identify $a = 5$ , $d = 3$ and $S_n=1025$ , then substitute:
$1025=n2[2(5)+(n−1)3]1025=\frac{n}{2}\left[2(5)+(n-1)3\right]$
Simplify to form a quadratic equation:
$2050=n(3n+7)3n2+7n−2050=0.\begin{aligned} 2050&=n(3n+7)\\ 3n^2+7n-2050&=0. \end{aligned}$
Factorise:
$(3 n + 82) (n - 25) = 0$
The solutions are $n=−823n=-\frac{82}{3}$ and $n = 25$ . Reject the negative solution because a number of terms must be a positive integer, so 25 terms are included.

Modelling with arithmetic sequences

Arithmetic sequences can model quantities that change by a fixed amount in equal stages, such as annual payments increasing by the same amount each year.

Before using the formulas, decide what one term represents and match the first stage to $n = 1$ .

Example

Modelling increasing annual payments

A payment is £600 in the first year and rises by £45 each year. Find the payment in year 10 and the total paid over the first 10 years.

Model the yearly payments as an arithmetic sequence with $a = 600$ , $d = 45$ and $n = 10$ .
Find the tenth payment:
$u_{10}=600+(10-1)45=1005$
The payment in year 10 is £1005.
Find the total using the first and last payments:
$S10=102(600+1005)=8025S_{10}=\frac{10}{2}(600+1005)=8025$
The total paid is £8025.

Exam technique

In the exam

Write down $a$ , $d$ and $n$ before substituting, paying particular attention to the sign of $d$ .
Use $u_n=a+(n-1)d$ for an individual term and a formula for $S_n$ only when terms are being added.
Check that any calculated term position or number of terms is a positive integer, and interpret the answer in its context.
Keep exact values throughout unless the question explicitly asks for a decimal approximation.

Self review

Check yourself

How would you find the common difference and nth term of $11,…19,\ 15,\ 11,\ldots$ ?
An arithmetic sequence has $u_4=17$ and $u_{10}=41$ . Can you find its first term and common difference?
Which form of the sum formula would be most efficient if you knew the first term, last term and number of terms?

What you'll learn

Sequences and series

Arithmetic sequences

Finding the nth term

Writing the nth term in a simpler form

Finding the position of a term

Using information about different terms

Arithmetic series

Deriving the sum formula

Finding the number of terms from a sum

Modelling with arithmetic sequences

Review 1.4.8 Arithmetic sequences and series (A-level only) by teaching Genie

1.4.8 Arithmetic sequences and series (A-level only)

What you'll learn

Sequences and series

Arithmetic sequences

Finding the nth term

Writing the nth term in a simpler form

Finding the position of a term

Using information about different terms

Arithmetic series

Deriving the sum formula

Finding the number of terms from a sum

Modelling with arithmetic sequences

Review 1.4.8 Arithmetic sequences and series (A-level only) by teaching Genie

1.4.8 Arithmetic sequences and series (A-level only)

What you'll learn

Sequences and series

Arithmetic sequences

Finding the nth term

Writing the nth term in a simpler form

Finding the position of a term

Using information about different terms

Arithmetic series

Deriving the sum formula

Finding the number of terms from a sum

Modelling with arithmetic sequences

Review 1.4.8 Arithmetic sequences and series (A-level only) by teaching Genie

1.4.8 Arithmetic sequences and series (A-level only) Revision Guide

1.4.8 Arithmetic sequences and series (A-level only)

What you'll learn

Sequences and series

Arithmetic sequences

Finding the nth term

Writing the nth term in a simpler form

Finding the position of a term

Using information about different terms

Arithmetic series

Deriving the sum formula

Finding the number of terms from a sum

Modelling with arithmetic sequences

Review 1.4.8 Arithmetic sequences and series (A-level only) by teaching Genie

1.4.8 Arithmetic sequences and series (A-level only) Revision Guide