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

What you'll learn

How to recognise an arithmetic sequence and find its common difference.
How to use and derive the formula for the $n$ th term.
How to distinguish a sequence from a series.
How to calculate the sum of the first $n$ terms of an arithmetic sequence.

Sequences and terms

A sequence is an ordered list of numbers. Each number in the sequence is called a term.

For example, in the sequence

…5,\ 8,\ 11,\ 14,\ \ldots

the first term is 5, the second term is 8, and so on. The dots show that the sequence continues.

We usually write the first term as $a$ . The term in position $n$ is called the $n$ th term and may be written as $u_n$ .

Definition

The nth term

The $n$ th term, $u_n$ , is a formula that gives the value of the term in position $n$ .

Arithmetic sequences

An arithmetic sequence is a sequence in which the difference between consecutive terms is constant.

This constant amount is called the common difference, written as $d$ .

For example,

…4,\ 9,\ 14,\ 19,\ \ldots

is arithmetic because each term is found by adding 5. Therefore, $a = 4$ and $d = 5$ .

The common difference can be negative. For example,

…20,\ 17,\ 14,\ 11,\ \ldots

has common difference $d = - 3$ .

Key Idea

Recognising an arithmetic sequence

Subtract each term from the term immediately after it. If the difference is always the same, the sequence is arithmetic.

Example

Finding the common difference

Determine whether the sequence $…31,\ 24,\ 17,\ 10,\ \ldots$ is arithmetic.

Calculate the differences between consecutive terms:
$24−31=−7,17−24=−7,10−17=−724-31=-7,\qquad 17-24=-7,\qquad 10-17=-7$
The differences are all equal, so the sequence is arithmetic.
Its first term is $a = 31$ and its common difference is $d = - 7$ .

Common Mistake

Reversing the subtraction

Calculate later term minus earlier term. For example, the common difference above is $24 - 31 = - 7$ , not $31 - 24 = 7$ .

Finding the nth term

Starting from the first term $a$ :

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

The multiplier of $d$ is always one less than the term number. Therefore,

u_n=a+(n-1)d

Definition

Nth term of an arithmetic sequence

For an arithmetic sequence with first term $a$ and common difference $d$ ,

u_n=a+(n-1)d

Example

Finding a specified term

Find the 40th term of the arithmetic sequence $…7,\ 12,\ 17,\ 22,\ \ldots$ .

Identify the first term and common difference:
$a=7,d=5a=7,\qquad d=5$
Substitute $n = 40$ into the formula:
$u_{40}=7+(40-1)(5)$
Calculate the term:
$u_{40}=7+195=202$

Common Mistake

Using n instead of n minus 1

The first term has had the common difference added zero times. This is why the formula contains $n - 1$ , not $n$ .

Finding the position of a term

You can also use the $n$ th-term formula to determine whether a particular value appears in a sequence. Substitute the value for $u_n$ and solve the resulting equation for $n$ .

A value is a term of the sequence only if the solution for $n$ is a positive integer.

Example

Checking whether a value is in a sequence

Determine whether 250 is a term of the arithmetic sequence $…10,\ 16,\ 22,\ 28,\ \ldots$ .

Here, $a = 10$ and $d = 6$ , so set $u_n=250$ :
$250 = 10 + (n - 1) 6$
Solve for $n$ :
$240=6(n−1)40=n−1n=41\begin{aligned} 240&=6(n-1)\\ 40&=n-1\\ n&=41 \end{aligned}$
Since 41 is a positive integer, 250 is the 41st term of the sequence.

Tip

Check the value of n

If solving for $n$ gives a decimal, a fraction, zero or a negative value, the proposed number is not a term of the sequence.

Finding an arithmetic sequence from given terms

Sometimes you are given two terms that are not consecutive. In that case, write an equation for each term using

u_n=a+(n-1)d

You can then solve the equations simultaneously to find $a$ and $d$ .

Example

Finding the first term and common difference

In an arithmetic sequence, $u_5=18$ and $u_{12}=46$ . Find $a$ and $d$ .

Use the formula for each given term:
$a+4d=18a+11d=46\begin{aligned} a+4d&=18\\ a+11d&=46 \end{aligned}$
Subtract the first equation from the second to eliminate $a$ :
$7 d = 28$
Therefore, $d = 4$ .
Substitute $d = 4$ into $a + 4 d = 18$ :
$a + 16 = 18$
Hence, $a = 2$ .
The sequence begins $…2,\ 6,\ 10,\ 14,\ \ldots$ .

From sequences to series

A series is formed when the terms of a sequence are added together.

For example,

5+8+11+14+⋯5+8+11+14+\cdots

is the arithmetic series associated with the arithmetic sequence $14,…5,\ 8,\ 11,\ 14,\ldots$

A finite series has a fixed number of terms. We write $S_n$ for the sum of the first $n$ terms.

Definition

Arithmetic series

An arithmetic series is the sum of the terms of an arithmetic sequence. The symbol $S_n$ represents the sum of its first $n$ terms.

Common Mistake

Confusing a sequence with a series

A sequence is a list of terms, separated by commas. A series is a sum of terms, joined by addition signs.

Deriving the sum formula

Suppose an arithmetic series has first term $a$ , common difference $d$ and $n$ terms:

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

Write the same sum in reverse:

Sn=(a+(n−1)d)+(a+(n−2)d)+⋯+aS_n=\left(a+(n-1)d\right)+\left(a+(n-2)d\right)+\cdots+a

Adding the two lines term by term gives $n$ identical pairs, each equal to $2 a + (n - 1) d$ :

2Sn=n(2a+(n−1)d)2S_n=n\left(2a+(n-1)d\right)

Therefore,

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

If the final term is called $l$ , then $l = a + (n - 1) d$ . This gives an alternative form:

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

Key Idea

Two forms of the sum formula

Use

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

when you know $a$ , $d$ and $n$ . Use

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

when you know the first term, last term and number of terms.

Example

Summing an arithmetic series

Find the sum of the first 30 terms of the arithmetic sequence $23,…8,\ 13,\ 18,\ 23,\ldots$

Identify the required values:
$a=8,d=5,n=30a=8,\qquad d=5,\qquad n=30$
Substitute into the sum formula:
$S30=302(2(8)+(30−1)5)S_{30}=\frac{30}{2}\left(2(8)+(30-1)5\right)$
Simplify:
$S30=15(16+145)=15(161)=2415\begin{aligned} S_{30}&=15(16+145)\\ &=15(161)\\ &=2415 \end{aligned}$

Finding the number of terms

If a sum is given, substituting into the formula for $S_n$ may produce a quadratic equation in $n$ .

You must solve the quadratic and then decide which solution is valid. Since $n$ is a number of terms, it must be a positive integer.

Example

Finding how many terms were added

The arithmetic series $3+7+11+⋯3+7+11+\cdots$ has sum 528. Find the number of terms.

Identify $a = 3$ , $d = 4$ and $S_n=528$ , then substitute:
$528=n2(2(3)+(n−1)4)528=\frac{n}{2}\left(2(3)+(n-1)4\right)$
Simplify to form a quadratic equation:
$528=n2(4n+2)528=n(2n+1)2n2+n−528=0\begin{aligned} 528&=\frac{n}{2}(4n+2)\\ 528&=n(2n+1)\\ 2n^2+n-528&=0 \end{aligned}$
Factorise:
$(2 n + 33) (n - 16) = 0$
Therefore, $n=−332n=-\frac{33}{2}$ or $n = 16$ .
Reject the negative solution because a series cannot contain a negative number of terms. The series has 16 terms.

Common Mistake

Not every stated sum is possible

If the equation for $n$ has no positive integer solution, then the given value cannot be the sum of the first $n$ terms of that arithmetic series.

Sigma notation

The Greek capital letter sigma, $∑\sum$ , means “add”. For example,

∑r=1n(a+(r−1)d)\sum_{r=1}^{n}\left(a+(r-1)d\right)

means add the terms $a + (r - 1) d$ as $r$ takes the integer values from 1 to $n$ . This is another way of writing $S_n$ .

Here, $r$ is an index: it keeps track of which term is being added.

Example

Evaluating a sum in sigma notation

Evaluate

∑r=120(3r+2)\sum_{r=1}^{20}(3r+2)

The expression $3 r + 2$ generates an arithmetic sequence. Its first term is obtained using $r = 1$ :
$a = 3 (1) + 2 = 5$
The coefficient of $r$ gives the common difference, so $d = 3$ , and there are 20 terms.
Apply the sum formula:
$S20=202(2(5)+(20−1)3)=10(10+57)=670\begin{aligned} S_{20}&=\frac{20}{2}\left(2(5)+(20-1)3\right)\\ &=10(10+57)\\ &=670 \end{aligned}$

Exam technique

In the exam

Write down $a$ , $d$ and $n$ before substituting into a formula, paying particular attention to a negative common difference.
Decide whether the question asks for one term, requiring $u_n=a+(n-1)d$ , or a total, requiring a formula for $S_n$ .
When solving for the number of terms, reject any solution that is not a positive integer.
Check that your answer is sensible: terms should change by $d$ , and a sum should reflect the approximate size and number of its terms.

Self review

Check yourself

What is the 25th term of the arithmetic sequence $4,…14,\ 9,\ 4,\ldots$ ?
An arithmetic sequence has $u_4=19$ and $u_{10}=43$ . Can you find its first term and common difference?
How many terms of the series $5+8+11+⋯5+8+11+\cdots$ are needed to give a sum of 1850?

What you'll learn

How to recognise an arithmetic sequence and find its common difference.
How to use and derive the formula for the $n$ th term.
How to distinguish a sequence from a series.
How to calculate the sum of the first $n$ terms of an arithmetic sequence.

Sequences and terms

A sequence is an ordered list of numbers. Each number in the sequence is called a term.

For example, in the sequence

…5,\ 8,\ 11,\ 14,\ \ldots

the first term is 5, the second term is 8, and so on. The dots show that the sequence continues.

We usually write the first term as $a$ . The term in position $n$ is called the $n$ th term and may be written as $u_n$ .

Definition

The nth term

The $n$ th term, $u_n$ , is a formula that gives the value of the term in position $n$ .

Arithmetic sequences

An arithmetic sequence is a sequence in which the difference between consecutive terms is constant.

This constant amount is called the common difference, written as $d$ .

For example,

…4,\ 9,\ 14,\ 19,\ \ldots

is arithmetic because each term is found by adding 5. Therefore, $a = 4$ and $d = 5$ .

The common difference can be negative. For example,

…20,\ 17,\ 14,\ 11,\ \ldots

has common difference $d = - 3$ .

Key Idea

Recognising an arithmetic sequence

Subtract each term from the term immediately after it. If the difference is always the same, the sequence is arithmetic.

Example

Finding the common difference

Determine whether the sequence $…31,\ 24,\ 17,\ 10,\ \ldots$ is arithmetic.

Calculate the differences between consecutive terms:
$24−31=−7,17−24=−7,10−17=−724-31=-7,\qquad 17-24=-7,\qquad 10-17=-7$
The differences are all equal, so the sequence is arithmetic.
Its first term is $a = 31$ and its common difference is $d = - 7$ .

Common Mistake

Reversing the subtraction

Calculate later term minus earlier term. For example, the common difference above is $24 - 31 = - 7$ , not $31 - 24 = 7$ .

Finding the nth term

Starting from the first term $a$ :

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

The multiplier of $d$ is always one less than the term number. Therefore,

u_n=a+(n-1)d

Definition

Nth term of an arithmetic sequence

For an arithmetic sequence with first term $a$ and common difference $d$ ,

u_n=a+(n-1)d

Example

Finding a specified term

Find the 40th term of the arithmetic sequence $…7,\ 12,\ 17,\ 22,\ \ldots$ .

Identify the first term and common difference:
$a=7,d=5a=7,\qquad d=5$
Substitute $n = 40$ into the formula:
$u_{40}=7+(40-1)(5)$
Calculate the term:
$u_{40}=7+195=202$

Common Mistake

Using n instead of n minus 1

The first term has had the common difference added zero times. This is why the formula contains $n - 1$ , not $n$ .

Finding the position of a term

You can also use the $n$ th-term formula to determine whether a particular value appears in a sequence. Substitute the value for $u_n$ and solve the resulting equation for $n$ .

A value is a term of the sequence only if the solution for $n$ is a positive integer.

Example

Checking whether a value is in a sequence

Determine whether 250 is a term of the arithmetic sequence $…10,\ 16,\ 22,\ 28,\ \ldots$ .

Here, $a = 10$ and $d = 6$ , so set $u_n=250$ :
$250 = 10 + (n - 1) 6$
Solve for $n$ :
$240=6(n−1)40=n−1n=41\begin{aligned} 240&=6(n-1)\\ 40&=n-1\\ n&=41 \end{aligned}$
Since 41 is a positive integer, 250 is the 41st term of the sequence.

Tip

Check the value of n

If solving for $n$ gives a decimal, a fraction, zero or a negative value, the proposed number is not a term of the sequence.

Finding an arithmetic sequence from given terms

Sometimes you are given two terms that are not consecutive. In that case, write an equation for each term using

u_n=a+(n-1)d

You can then solve the equations simultaneously to find $a$ and $d$ .

Example

Finding the first term and common difference

In an arithmetic sequence, $u_5=18$ and $u_{12}=46$ . Find $a$ and $d$ .

Use the formula for each given term:
$a+4d=18a+11d=46\begin{aligned} a+4d&=18\\ a+11d&=46 \end{aligned}$
Subtract the first equation from the second to eliminate $a$ :
$7 d = 28$
Therefore, $d = 4$ .
Substitute $d = 4$ into $a + 4 d = 18$ :
$a + 16 = 18$
Hence, $a = 2$ .
The sequence begins $…2,\ 6,\ 10,\ 14,\ \ldots$ .

From sequences to series

A series is formed when the terms of a sequence are added together.

For example,

5+8+11+14+⋯5+8+11+14+\cdots

is the arithmetic series associated with the arithmetic sequence $14,…5,\ 8,\ 11,\ 14,\ldots$

A finite series has a fixed number of terms. We write $S_n$ for the sum of the first $n$ terms.

Definition

Arithmetic series

An arithmetic series is the sum of the terms of an arithmetic sequence. The symbol $S_n$ represents the sum of its first $n$ terms.

Common Mistake

Confusing a sequence with a series

A sequence is a list of terms, separated by commas. A series is a sum of terms, joined by addition signs.

Deriving the sum formula

Suppose an arithmetic series has first term $a$ , common difference $d$ and $n$ terms:

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

Write the same sum in reverse:

Sn=(a+(n−1)d)+(a+(n−2)d)+⋯+aS_n=\left(a+(n-1)d\right)+\left(a+(n-2)d\right)+\cdots+a

Adding the two lines term by term gives $n$ identical pairs, each equal to $2 a + (n - 1) d$ :

2Sn=n(2a+(n−1)d)2S_n=n\left(2a+(n-1)d\right)

Therefore,

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

If the final term is called $l$ , then $l = a + (n - 1) d$ . This gives an alternative form:

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

Key Idea

Two forms of the sum formula

Use

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

when you know $a$ , $d$ and $n$ . Use

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

when you know the first term, last term and number of terms.

Example

Summing an arithmetic series

Find the sum of the first 30 terms of the arithmetic sequence $23,…8,\ 13,\ 18,\ 23,\ldots$

Identify the required values:
$a=8,d=5,n=30a=8,\qquad d=5,\qquad n=30$
Substitute into the sum formula:
$S30=302(2(8)+(30−1)5)S_{30}=\frac{30}{2}\left(2(8)+(30-1)5\right)$
Simplify:
$S30=15(16+145)=15(161)=2415\begin{aligned} S_{30}&=15(16+145)\\ &=15(161)\\ &=2415 \end{aligned}$

Finding the number of terms

If a sum is given, substituting into the formula for $S_n$ may produce a quadratic equation in $n$ .

You must solve the quadratic and then decide which solution is valid. Since $n$ is a number of terms, it must be a positive integer.

Example

Finding how many terms were added

The arithmetic series $3+7+11+⋯3+7+11+\cdots$ has sum 528. Find the number of terms.

Identify $a = 3$ , $d = 4$ and $S_n=528$ , then substitute:
$528=n2(2(3)+(n−1)4)528=\frac{n}{2}\left(2(3)+(n-1)4\right)$
Simplify to form a quadratic equation:
$528=n2(4n+2)528=n(2n+1)2n2+n−528=0\begin{aligned} 528&=\frac{n}{2}(4n+2)\\ 528&=n(2n+1)\\ 2n^2+n-528&=0 \end{aligned}$
Factorise:
$(2 n + 33) (n - 16) = 0$
Therefore, $n=−332n=-\frac{33}{2}$ or $n = 16$ .
Reject the negative solution because a series cannot contain a negative number of terms. The series has 16 terms.

Common Mistake

Not every stated sum is possible

If the equation for $n$ has no positive integer solution, then the given value cannot be the sum of the first $n$ terms of that arithmetic series.

Sigma notation

The Greek capital letter sigma, $∑\sum$ , means “add”. For example,

∑r=1n(a+(r−1)d)\sum_{r=1}^{n}\left(a+(r-1)d\right)

means add the terms $a + (r - 1) d$ as $r$ takes the integer values from 1 to $n$ . This is another way of writing $S_n$ .

Here, $r$ is an index: it keeps track of which term is being added.

Example

Evaluating a sum in sigma notation

Evaluate

∑r=120(3r+2)\sum_{r=1}^{20}(3r+2)

The expression $3 r + 2$ generates an arithmetic sequence. Its first term is obtained using $r = 1$ :
$a = 3 (1) + 2 = 5$
The coefficient of $r$ gives the common difference, so $d = 3$ , and there are 20 terms.
Apply the sum formula:
$S20=202(2(5)+(20−1)3)=10(10+57)=670\begin{aligned} S_{20}&=\frac{20}{2}\left(2(5)+(20-1)3\right)\\ &=10(10+57)\\ &=670 \end{aligned}$

Exam technique

In the exam

Write down $a$ , $d$ and $n$ before substituting into a formula, paying particular attention to a negative common difference.
Decide whether the question asks for one term, requiring $u_n=a+(n-1)d$ , or a total, requiring a formula for $S_n$ .
When solving for the number of terms, reject any solution that is not a positive integer.
Check that your answer is sensible: terms should change by $d$ , and a sum should reflect the approximate size and number of its terms.

Self review

Check yourself

What is the 25th term of the arithmetic sequence $4,…14,\ 9,\ 4,\ldots$ ?
An arithmetic sequence has $u_4=19$ and $u_{10}=43$ . Can you find its first term and common difference?
How many terms of the series $5+8+11+⋯5+8+11+\cdots$ are needed to give a sum of 1850?

1.7.4 Arithmetic sequences and series (A-level only)

What you'll learn

Sequences and terms

Arithmetic sequences

Finding the nth term

Finding the position of a term

Finding an arithmetic sequence from given terms

From sequences to series

Deriving the sum formula

Finding the number of terms

Sigma notation

Review 1.6.12 Arithmetic sequences and series (A-level only) by teaching Genie

1.7.4 Arithmetic sequences and series (A-level only)

What you'll learn

Sequences and terms

Arithmetic sequences

Finding the nth term

Finding the position of a term

Finding an arithmetic sequence from given terms

From sequences to series

Deriving the sum formula

Finding the number of terms

Sigma notation

Review 1.6.12 Arithmetic sequences and series (A-level only) by teaching Genie

1.7.4 Arithmetic sequences and series (A-level only)

What you'll learn

Sequences and terms

Arithmetic sequences

Finding the nth term

Finding the position of a term

Finding an arithmetic sequence from given terms

From sequences to series

Deriving the sum formula

Finding the number of terms

Sigma notation

Review 1.6.12 Arithmetic sequences and series (A-level only) by teaching Genie

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

1.7.4 Arithmetic sequences and series (A-level only)

What you'll learn

Sequences and terms

Arithmetic sequences

Finding the nth term

Finding the position of a term

Finding an arithmetic sequence from given terms

From sequences to series

Deriving the sum formula

Finding the number of terms

Sigma notation

Review 1.6.12 Arithmetic sequences and series (A-level only) by teaching Genie

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