GCSE Maths revision · Higher: number and algebra

Quadratic sequences

Higher only

The idea

In a quadratic sequence the second differences are constant. The $n$ th term is $an^2 + bn + c$ , where $a$ is half the second difference.

Find the second difference and halve it to get $a$ .
Subtract $an^2$ from each term; what is left is a linear sequence.
Find its $n$ th term, $bn + c$ , and add it on.

Watch out The second difference is $2a$ , not $a$ .

See it worked

Worked exampleNon-calculator3 marks

Here are the first five terms of a quadratic sequence: $3, 8, 15, 24, 35$ .

Find an expression for the $n$ th term.

Differences: 5, 7, 9, 11. Second difference: 2, so the $n^2$ coefficient is $2 \div 2 = 1$ . $n^2: \ 1, 4, 9, 16, 25$
Subtract $n^2$ from each term. $3 - 1 = 2, \ 8 - 4 = 4, \ 15 - 9 = 6, \ \ldots \Rightarrow 2n$
So the $n$ th term is $n^2 + 2n$ .

Answer $n^2 + 2n$

Your turn

Question 1Non-calculator3 marks

The first four terms of a quadratic sequence are 6, 13, 24 and 39. Find an expression for its nth term, starting at n = 1.

Question 2Non-calculator4 marks

The nth term of a sequence is $2n^2 + 3n - 1$ .

(a) Find the 5th term.

(b) Which term of the sequence is 229?

Question 3Calculator allowed5 marks

A quadratic sequence has constant second difference 2. Its second term is 9 and its fifth term is 42. Starting with the first term at n = 1, find the position of the first term greater than 1000.

In past papers

20 official questions in 20 papers since 2017. The latest:

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