Arithmetic Sequences

An arithmetic sequence (also called an arithmetic progression, or AP) is a sequence where each term is obtained by adding a fixed number, called the common difference, to the previous term. This topic is fundamental to A-Level Mathematics (9MA0) and appears across Papers 1 and 2.

Key Definitions

Term	Meaning
Arithmetic sequence	A sequence with a constant difference between consecutive terms
Common difference (d)	The fixed value added to each term: d = u(n+1) - u(n)
First term (a)	The starting value of the sequence, u(1)
nth term, u(n)	The general term of the sequence
Arithmetic series	The sum of the terms of an arithmetic sequence

The nth Term Formula

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

u(n) = a + (n - 1)d

This is given in the Edexcel formula booklet but you must be able to use it fluently.

Worked Example 1

Find the 20th term of the arithmetic sequence 5, 8, 11, 14, ...

First term: a = 5
Common difference: d = 8 - 5 = 3
u(20) = 5 + (20 - 1) x 3 = 5 + 57 = 62

Worked Example 2

The 3rd term of an AP is 10 and the 7th term is 22. Find a and d.

Using u(n) = a + (n - 1)d:

u(3) = a + 2d = 10 ... (1)
u(7) = a + 6d = 22 ... (2)

Subtract (1) from (2): 4d = 12, so d = 3

Substitute back: a + 6 = 10, so a = 4

Answer: a = 4, d = 3

Worked Example 3

The nth term of an arithmetic sequence is u(n) = 7n - 2. State the first term and common difference.

u(1) = 7(1) - 2 = 5, so a = 5
u(2) = 7(2) - 2 = 12
d = 12 - 5 = 7

Alternatively, comparing u(n) = 7n - 2 with u(n) = a + (n-1)d = dn + (a - d):

Coefficient of n gives d = 7
Constant gives a - d = -2, so a = 5

Sum of n Terms

The sum of the first n terms of an arithmetic series is:

S(n) = n/2 x (2a + (n - 1)d)

or equivalently:

S(n) = n/2 x (a + l)

where l is the last term.

Both forms are in the formula booklet. Choose whichever is more convenient for the given information.

Derivation (you should understand this)

Write the sum forwards and backwards:

S(n) = a + (a + d) + (a + 2d) + ... + (a + (n-1)d) S(n) = (a + (n-1)d) + (a + (n-2)d) + ... + a

Adding these two equations term by term:

2 x S(n) = n x (2a + (n-1)d)

So S(n) = n/2 x (2a + (n-1)d)

Worked Example 4

Find the sum of the first 50 terms of the arithmetic series 3 + 7 + 11 + 15 + ...

a = 3, d = 4, n = 50
S(50) = 50/2 x (2(3) + 49(4))
S(50) = 25 x (6 + 196)
S(50) = 25 x 202 = 5050

Worked Example 5

The sum of the first 20 terms of an AP is 610 and the first term is 4. Find d.

S(20) = 20/2 x (2(4) + 19d) = 610
10 x (8 + 19d) = 610
8 + 19d = 61
19d = 53
d = 53/19

Answer: d = 53/19

Finding the Number of Terms

Sometimes you need to find how many terms are in a given range or how many terms are needed for the sum to exceed a value.

Worked Example 6

How many terms of the AP 2, 5, 8, 11, ... are needed for the sum to first exceed 1000?

S(n) = n/2 x (2(2) + (n-1)(3)) = n/2 x (4 + 3n - 3) = n/2 x (3n + 1)

We need S(n) > 1000:

n(3n + 1)/2 > 1000

3n² + n > 2000

3n² + n - 2000 > 0

Using the quadratic formula: n = (-1 + sqrt(1 + 24000)) / 6 = (-1 + sqrt(24001)) / 6

sqrt(24001) ≈ 154.93, so n > (-1 + 154.93)/6 ≈ 25.65

Since n must be a positive integer, n = 26 terms are needed.

Check: S(25) = 25/2 x (76) = 950, S(26) = 26/2 x (79) = 1027. Confirmed.

Real-World Applications

Worked Example 7

A trainee saves £100 in the first month and increases the amount by £15 each month. How much has she saved in total after 2 years?

a = 100, d = 15, n = 24 (months)
S(24) = 24/2 x (2(100) + 23(15))
S(24) = 12 x (200 + 345)
S(24) = 12 x 545 = £6540

Worked Example 8

Seats in a theatre are arranged so that the first row has 20 seats and each subsequent row has 2 more seats than the row in front. If there are 30 rows, how many seats are there in total?

a = 20, d = 2, n = 30
S(30) = 30/2 x (2(20) + 29(2)) = 15 x (40 + 58) = 15 x 98 = 1470 seats

Exam Tips

Tip	Detail
Check your d	Always verify d by checking consecutive terms; a sign error is easy to make
Integer answers	When finding n (number of terms), the answer must be a positive integer — round UP if the sum must exceed a value
Two unknowns	If given two pieces of information, set up two equations using u(n) and/or S(n) and solve simultaneously
Notation	Edexcel uses both u(n) and a + (n-1)d interchangeably; be comfortable with both
Show working	Even with a formula booklet, examiners want to see substitution and simplification steps

Practice Problems

An AP has first term 8 and common difference -3. Find the 15th term. [Answer: -34]
The 5th term of an AP is 19 and the 10th term is 39. Find the first term and common difference. [Answer: a = 3, d = 4]
Find the sum of the first 100 positive integers. [Answer: 5050]
How many terms of 1 + 4 + 7 + 10 + ... give a sum of 1045? [Answer: 26]
A carpenter lays 15 bricks in the first hour, 18 in the second, 21 in the third, and so on. How many bricks in total after 8 hours? [Answer: 204]

Summary

An arithmetic sequence has constant common difference d.
nth term: u(n) = a + (n - 1)d
Sum of n terms: S(n) = n/2 x (2a + (n-1)d) = n/2 x (a + l)
To find n from a sum condition, set up an inequality and solve the resulting quadratic.
These formulae are in the booklet but must be applied fluently under exam conditions.