Cambridge IGCSE Mathematics 0580

Sequences and the nth term

The $n$ th term of a sequence is a formula that gives any term from its position $n$ . For a linear sequence — constant first difference $d$ — it is $d n + c$ ; for a quadratic sequence — constant second difference — it is $a n^{2} + b n + c$ with $a$ equal to half the second difference; for a geometric sequence — constant ratio $r$ — it is $a r^{n - 1}$ . Cambridge IGCSE Maths 0580 sets sequences on 21 of the 42 papers Quanta has mapped, for 113 marks, and the quadratic $n$ th term is where Extended candidates gain or lose the grade-8 marks.

Updated 15 September 2026

How much it is worth

Sequences and the nth term: 113 marks across 42 real Cambridge 0580 papers — 2.7% of the total, on 21 of the 42. How that ranks against every other 0580 topic →

Linear sequences

If the terms go up (or down) by the same amount each time, that amount is the coefficient of $n$ . Then adjust:

7, 11, 15, 19, \dots d = 4 \Rightarrow 4 n + c; n = 1 : 4 + c = 7 \Rightarrow c = 3 \Rightarrow 4 n + 3

The shortcut is $c = first term - d$ , which is the “zeroth term”. A decreasing sequence has a negative $d$ : $20, 17, 14, \dots$ is $- 3 n + 23$ . Always check with $n = 2$ as well as $n = 1$ ; a wrong sign in $c$ passes one check and fails the other.

Quadratic sequences

If the first differences are not constant but the second differences are, the sequence is quadratic and the method has three steps:

Find the second difference. $a$ is half of it.
Subtract $a n^{2}$ from each term. What remains is a linear sequence.
Find that linear sequence’s $n$ th term, $b n + c$ , and add it back.

The worked example below does this in full. Extended papers give these as a table of positions and terms, or as a pattern of diagrams whose counts form the sequence.

Cubic and geometric sequences

A constant third difference means a cubic term; the coefficient of $n^{3}$ is the third difference divided by 6. These are rare and usually structured — the question gives the form $a n^{3} + b$ and asks for $a$ and $b$ — so substitute two positions and solve simultaneously rather than differencing.

A geometric sequence multiplies by the same ratio each time:

3, 6, 12, 24, \dots r = 2 \Rightarrow 3 \times 2^{n - 1}

The power is $n - 1$ , not $n$ : the first term has been multiplied by $r$ zero times. Sequences of powers such as $2, 4, 8, 16$ ( $2^{n}$ ) or $1, 4, 9, 16$ ( $n^{2}$ ) are the special cases to recognise on sight.

Worked example

Find the nth term of the sequence 2, 9, 20, 35, 54, …

Differences: first differences $7, 11, 15, 19$ ; second differences $4, 4, 4$ — constant, so the sequence is quadratic with $a = 4 \div 2 = 2$ .

Subtract $2 n^{2}$ : $2 n^{2}$ gives $2, 8, 18, 32, 50$ . Term minus this:

2 - 2, 9 - 8, 20 - 18, 35 - 32, 54 - 50 = 0, 1, 2, 3, 4

Linear remainder: $0, 1, 2, 3, 4$ has $d = 1$ and first term 0, so it is $n - 1$ .

n th term = 2 n^{2} + n - 1

Check with a term not used in the working: $n = 5$ gives $50 + 5 - 1 = 54$ . ✓

A typical scheme gives a mark for the second difference or for $2 n^{2}$ , a mark for the linear remainder, and a mark for the final expression — so the method shown is the marks, even when the final line is wrong.

Using the nth term

Find a term: substitute. The 50th term of $4 n + 3$ is $203$ .
Is 275 a term? Solve $4 n + 3 = 275$ : $n = 68$ , a positive integer, so yes — the 68th. If $n$ is not a whole number, it is not a term, and that is the required reason.
First term above a value: solve the inequality, then round up to the next integer.
Which term is …? for a quadratic: solve the quadratic, keep the positive integer root.

Common mistakes

1.Giving the term-to-term rule as the nth term
“Add 4” describes how to get the next term. The $n$ th term is a formula in $n$ : $4 n + 3$ . Both may be asked; they are different answers.
2.Using the whole second difference as a
$a$ is half the second difference. A second difference of 4 gives $2 n^{2}$ , not $4 n^{2}$ .
3.Wrong sign on c
$c$ is what you add to $d n$ to reach the first term: $c = u_{1} - d$ . Check $n = 1$ and $n = 2$ before moving on.
4.Geometric nth term with power n
$3 \times 2^{n}$ starts at 6, not 3. The first term is multiplied by $r$ zero times, so the power is $n - 1$ .
5.Deciding 'is it a term' without solving
The reason the scheme wants is that solving for $n$ gives a non-integer. Show the equation and its solution.

Common questions

How do you find the nth term of a linear sequence?

The common difference $d$ is the coefficient of $n$ ; then $c = first term - d$ . So $5, 8, 11, \dots$ is $3 n + 2$ .

How do you find the nth term of a quadratic sequence?

Halve the constant second difference to get $a$ , subtract $a n^{2}$ from every term, find the $n$ th term of the linear sequence left over, and add the two together.

Are quadratic sequences on the Core paper?

Linear sequences and simple patterns (squares, cubes, powers) are Core; the general quadratic and cubic $n$ th term is Extended.

How do you show that a number is not in a sequence?

Set the $n$ th term equal to the number and solve for $n$ . If $n$ is not a positive whole number, the number is not a term — and that sentence is the explanation the scheme wants.

Sequences and the nth term on real papers

The papers that leaned on it hardest, as a share of the paper. Each one has its own page on Quanta with the full topic breakdown:

0580/22/F/M/237 of 70 marks 0580/43/M/J/2412 of 130 marks 0580/21/M/J/246 of 70 marks

Every 0580 past-paper question on sequences and the nth term

24 questions across 21 papers, 113 marks in all, newest first. Open a paper to sit it on Quanta and have these questions marked.

Paper	Question (marks)
0580/23/O/N/25	Q16 (5)
0580/41/O/N/25	Q20 (3)
0580/42/O/N/25	Q10 (4)
0580/43/O/N/25	Q1 (2)
0580/22/M/J/25	Q7 (3)
0580/41/M/J/25	Q5 (6)
0580/43/M/J/25	Q11 (8)
0580/42/F/M/25	Q13 (6)
0580/21/O/N/24	Q13 (5)
0580/22/O/N/24	Q1 (2), Q20 (2)
0580/23/O/N/24	Q13 (4)
0580/21/M/J/24	Q3 (4), Q8 (2)
0580/42/M/J/24	Q5 (2)
0580/43/M/J/24	Q2 (12)
0580/22/F/M/24	Q8 (4)
0580/22/O/N/23	Q19 (4)
0580/41/O/N/23	Q6 (11)
0580/21/M/J/23	Q8 (4)
0580/23/M/J/23	Q22 (5)
0580/42/M/J/23	Q6 (8)
0580/22/F/M/23	Q4 (3), Q20 (4)

Practise sequences and the nth term against real mark schemes

Quanta has real Cambridge IGCSE Maths 0580 past-paper questions, broken into skill checkpoints, marked criterion by criterion the way examiners mark — and it tracks which skills you’re missing. Free for individual students.

Start practising free