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How do you find the nth term of linear and quadratic sequences and recognise special sequences?

Generate sequences from a rule; find the nth term of a linear sequence and a quadratic sequence (Higher tier); and recognise arithmetic, geometric, square, cube, triangular and Fibonacci sequences.

A focused answer to the Eduqas GCSE Mathematics algebra content on sequences, covering generating terms, the nth term of linear and quadratic sequences, and recognising arithmetic, geometric, square, cube, triangular and Fibonacci sequences.

Generated by Claude Opus 4.810 min answer

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  1. What this dot point is asking
  2. Generating a sequence
  3. The nth term of a linear sequence
  4. The nth term of a quadratic sequence (Higher)
  5. Recognising special sequences
  6. Using the nth term
  7. Why sequences matter

What this dot point is asking

The Eduqas algebra content asks you to generate the terms of a sequence from a rule, find the nth term of a linear (arithmetic) sequence, and at Higher tier find the nth term of a quadratic sequence. You should also recognise common sequences on sight: arithmetic, geometric, square, cube, triangular and Fibonacci. The nth term is the key tool, because it lets you find any term directly and test whether a given number belongs to the sequence. Linear nth term appears at both tiers; the quadratic nth term is a reliable Higher-tier question.

Generating a sequence

A sequence can be defined by a position-to-term rule (the nth term) or a term-to-term rule (how to get from one term to the next). To generate from an nth term, substitute n=1,2,3,n = 1, 2, 3, \ldots: the rule 3n23n - 2 gives 1,4,7,10,1, 4, 7, 10, \ldots. To generate from a term-to-term rule, start with the first term and apply the rule repeatedly: "start at 55, double each time" gives 5,10,20,40,5, 10, 20, 40, \ldots.

The nth term of a linear sequence

A linear sequence increases by the same amount each time, the common difference dd.

For 5,8,11,14,5, 8, 11, 14, \ldots, the difference is 33, so the rule starts 3n3n. The 3n3n values are 3,6,9,123, 6, 9, 12, which are 22 short of the sequence, so c=2c = 2 and the nth term is 3n+23n + 2. To test whether 5050 is a term, solve 3n+2=503n + 2 = 50: this gives n=16n = 16, a whole number, so 5050 is the 16th term. If nn were not a whole number, the value would not be in the sequence.

The nth term of a quadratic sequence (Higher)

A quadratic sequence has a constant second difference (the differences of the differences).

Recognising special sequences

Some sequences should be known by sight. The square numbers are 1,4,9,16,1, 4, 9, 16, \ldots (nth term n2n^2); the cube numbers are 1,8,27,64,1, 8, 27, 64, \ldots (nth term n3n^3); the triangular numbers are 1,3,6,10,15,1, 3, 6, 10, 15, \ldots with nth term n(n+1)2\tfrac{n(n+1)}{2}. The Fibonacci sequence 1,1,2,3,5,8,1, 1, 2, 3, 5, 8, \ldots adds the previous two terms. A geometric sequence multiplies by a constant ratio each time, such as 2,6,18,54,2, 6, 18, 54, \ldots (ratio 33). Spotting which type a sequence is points you straight to the right method.

Using the nth term

The power of the nth term is that it gives any term without listing all the earlier ones. To find the 100th term of 3,7,11,3, 7, 11, \ldots with nth term 4n14n - 1, substitute n=100n = 100 to get 399399, with no need to write out the sequence. The nth term also tests membership: to ask whether 250250 is a term of 4n14n - 1, solve 4n1=2504n - 1 = 250, which gives n=62.75n = 62.75. Because nn is not a whole number, 250250 is not in the sequence. This "is it a term" question is a favourite Eduqas reasoning task because it rewards understanding that nn must be a positive integer.

Why sequences matter

Sequences link algebra to pattern-spotting and to real growth. A linear sequence models a quantity changing by a fixed amount each step (savings growing by the same deposit), while a geometric sequence models repeated proportional change (a population multiplying by a fixed factor), which connects directly to the growth and decay work in the ratio area. Finding and justifying an nth term is also an AO2 communication skill: Eduqas wants the rule stated clearly in terms of nn, not just the next few terms, so always give the general expression when one is asked for.

Exam-style practice questions

Practice questions written in the style of WJEC Eduqas exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Eduqas 20193 marksFind an expression, in terms of nn, for the nth term of the sequence 7,11,15,19,7, 11, 15, 19, \ldots and use it to find the 50th term. (Foundation, Component 1, non-calculator.)
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The first difference is constant at +4+4, so the sequence is linear with nth term 4n+c4n + c.

The 4n4n part gives 4,8,12,4, 8, 12, \ldots, which is 33 less than the sequence, so c=3c = 3 and the nth term is 4n+34n + 3.

The 50th term is 4×50+3=2034 \times 50 + 3 = 203.

Markers award a mark for 4n4n, a mark for the full rule 4n+34n + 3, and a mark for the 50th term. A frequent error is to write n+4n + 4 (treating the difference as an additive rule) instead of 4n+34n + 3.

Eduqas 20224 marksThe nth term of a quadratic sequence is required. The sequence is 3,8,15,24,35,3, 8, 15, 24, 35, \ldots. Find the nth term. (Higher, Component 1, non-calculator.)
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First differences are 5,7,9,115, 7, 9, 11; second differences are constant at 22.

The coefficient of n2n^2 is half the second difference: 22=1\dfrac{2}{2} = 1, so the n2n^2 part is n2n^2.

Subtract n2n^2 (1,4,9,16,251, 4, 9, 16, 25) from the sequence to leave 2,4,6,8,102, 4, 6, 8, 10, a linear sequence with nth term 2n2n.

So the nth term is n2+2nn^2 + 2n.

Markers give marks for the second difference, for the n2n^2 coefficient, for the linear remainder, and for the full rule. Forgetting to halve the second difference is the usual slip.

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