Cambridge IGCSE Mathematics 0580

Sequences and the nth term

The nnth term of a sequence is a formula that gives any term from its position nn. For a linear sequence — constant first difference dd — it is dn+cdn + c; for a quadratic sequence — constant second difference — it is an2+bn+can^2 + bn + c with aa equal to half the second difference; for a geometric sequence — constant ratio rr — it is arn1a r^{\,n-1}. Cambridge IGCSE Maths 0580 sets sequences on 21 of the 42 papers Quanta has mapped, for 113 marks, and the quadratic nnth term is where Extended candidates gain or lose the grade-8 marks.

Updated 15 September 2026

Linear sequences

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

7, 11, 15, 19, d=44n+c;  n=1:4+c=7c=34n+37,\ 11,\ 15,\ 19,\ \ldots \quad d = 4 \quad\Rightarrow\quad 4n + c;\ \ n=1: 4 + c = 7 \Rightarrow c = 3 \quad\Rightarrow\quad 4n + 3

The shortcut is c=first termdc = \text{first term} - d, which is the “zeroth term”. A decreasing sequence has a negative dd: 20,17,14,20, 17, 14, \ldots is 3n+23-3n + 23. Always check with n=2n = 2 as well as n=1n = 1; a wrong sign in cc 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:

  1. Find the second difference. aa is half of it.
  2. Subtract an2an^2 from each term. What remains is a linear sequence.
  3. Find that linear sequence’s nnth term, bn+cbn + 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 n3n^3 is the third difference divided by 6. These are rare and usually structured — the question gives the form an3+ban^3 + b and asks for aa and bb — so substitute two positions and solve simultaneously rather than differencing.

A geometric sequence multiplies by the same ratio each time:

3, 6, 12, 24, r=23×2n13,\ 6,\ 12,\ 24,\ \ldots \quad r = 2 \quad\Rightarrow\quad 3 \times 2^{\,n-1}

The power is n1n - 1, not nn: the first term has been multiplied by rr zero times. Sequences of powers such as 2,4,8,162, 4, 8, 16 (2n2^n) or 1,4,9,161, 4, 9, 16 (n2n^2) are the special cases to recognise on sight.

Worked example

Worked example

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

Differences: first differences 7,11,15,197, 11, 15, 19; second differences 4,4,44, 4, 4 — constant, so the sequence is quadratic with a=4÷2=2a = 4 \div 2 = 2.

Subtract 2n22n^2: 2n22n^2 gives 2,8,18,32,502, 8, 18, 32, 50. Term minus this:

22, 98, 2018, 3532, 5450  =  0, 1, 2, 3, 42-2,\ 9-8,\ 20-18,\ 35-32,\ 54-50 \;=\; 0,\ 1,\ 2,\ 3,\ 4

Linear remainder: 0,1,2,3,40, 1, 2, 3, 4 has d=1d = 1 and first term 0, so it is n1n - 1.

nth term=2n2+n1n\text{th term} = 2n^2 + n - 1

Check with a term not used in the working: n=5n = 5 gives 50+51=5450 + 5 - 1 = 54. ✓

A typical scheme gives a mark for the second difference or for 2n22n^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 4n+34n + 3 is 203203.
  • Is 275 a term? Solve 4n+3=2754n + 3 = 275: n=68n = 68, a positive integer, so yes — the 68th. If nn 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. 1.Giving the term-to-term rule as the nth term

    “Add 4” describes how to get the next term. The nnth term is a formula in nn: 4n+34n + 3. Both may be asked; they are different answers.

  2. 2.Using the whole second difference as a

    aa is half the second difference. A second difference of 4 gives 2n22n^2, not 4n24n^2.

  3. 3.Wrong sign on c

    cc is what you add to dndn to reach the first term: c=u1dc = u_1 - d. Check n=1n = 1 and n=2n = 2 before moving on.

  4. 4.Geometric nth term with power n

    3×2n3 \times 2^n starts at 6, not 3. The first term is multiplied by rr zero times, so the power is n1n - 1.

  5. 5.Deciding 'is it a term' without solving

    The reason the scheme wants is that solving for nn 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 dd is the coefficient of nn; then c=first termdc = \text{first term} - d. So 5,8,11,5, 8, 11, \ldots is 3n+23n + 2.

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

Halve the constant second difference to get aa, subtract an2an^2 from every term, find the nnth 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 nnth term is Extended.

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

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

Practise sequences and the nth term against real mark schemes

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