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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 OCR GCSE Mathematics algebra content on sequences, covering the nth term of linear and quadratic sequences, generating terms from a rule, and recognising special sequences.

Generated by Claude Opus 4.810 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this dot point is asking
  2. Linear sequences and the nth term
  3. Recognising special sequences
  4. Quadratic sequences (Higher)
  5. Why the nth term matters

What this dot point is asking

OCR reference A5 asks you to generate sequences from a term-to-term or position-to-term rule, to find the nth term of a linear sequence (and a quadratic sequence at Higher tier), and to recognise special sequences such as arithmetic, geometric, square, cube, triangular and Fibonacci. The nth term is the engine: it lets you find any term directly and test whether a value belongs. Sequences appear on every paper and link to straight-line and quadratic graphs.

Linear sequences and the nth term

A linear sequence goes up (or down) by the same amount each time.

For 4,7,10,134, 7, 10, 13 the difference is 33, so the rule starts 3n3n; since 3×1=33 \times 1 = 3 but the first term is 44, add 11 to get 3n+13n + 1. A decreasing sequence has a negative difference: 20,17,14,1120, 17, 14, 11 has nth term 3n+23-3n + 23. The nth term answers both "what is the 100100th term?" (substitute n=100n = 100) and "is 6161 in the sequence?" (solve 3n+1=613n + 1 = 61 and check nn is a whole number).

Recognising special sequences

Several named sequences recur in exams.

The square numbers are 1,4,9,16,251, 4, 9, 16, 25 (nth term n2n^2); the cube numbers are 1,8,27,641, 8, 27, 64 (nth term n3n^3); the triangular numbers are 1,3,6,10,151, 3, 6, 10, 15 (nth term n(n+1)2\tfrac{n(n+1)}{2}). A Fibonacci-type sequence adds the previous two terms, as in 2,3,5,8,132, 3, 5, 8, 13. A geometric sequence multiplies by a constant ratio, as in 3,6,12,243, 6, 12, 24 (ratio 22). Spotting these quickly often unlocks a question that looks harder than it is.

OCR sometimes describes a sequence by a position-to-term rule (substitute the position nn to get the term) or a term-to-term rule (do something to the previous term). A position-to-term rule such as n2+1n^2 + 1 lets you jump straight to the 2020th term, whereas a term-to-term rule such as "double and add one" forces you to work through each term in turn. Being asked to convert between the two descriptions, or to continue a sequence given one of them, is common, so be comfortable reading both.

Quadratic sequences (Higher)

A quadratic sequence has a changing first difference but a constant second difference.

Why the nth term matters

The nth term turns a list into a formula, which is exactly the position-to-term thinking OCR wants. It connects to graphs: a linear sequence plotted against nn gives points on a straight line, and a quadratic sequence gives points on a parabola. Forming and using the nth term is also an AO2 reasoning skill, so showing the difference table and the adjustment step secures method marks even if the final expression has a slip.

Exam-style practice questions

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

OCR 20193 marksThe first four terms of a sequence are 7,12,17,227, 12, 17, 22. Find an expression for the nth term, and use it to find the 5050th term. (Foundation, Paper 2, non-calculator.)
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The common difference is 55, so the nth term begins with 5n5n.

Compare 5n5n (5,10,15,205, 10, 15, 20) with the sequence (7,12,17,227, 12, 17, 22): each term is 22 more, so the nth term is 5n+25n + 2.

The 5050th term is 5(50)+2=2525(50) + 2 = 252.

Markers award a mark for the 5n5n part, a mark for the full nth term 5n+25n + 2, and a mark for the 5050th term 252252. Writing the rule as "add 55" rather than an nth-term expression scores nothing for the first marks.

OCR 20224 marksFind the nth term of the quadratic sequence 3,8,15,24,353, 8, 15, 24, 35. (Higher, Paper 5, non-calculator.)
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First differences are 5,7,9,115, 7, 9, 11; second differences are constant at 22.

The n2n^2 coefficient is half the second difference: 2÷2=12 \div 2 = 1, so start with n2n^2.

Subtract n2n^2 (1,4,9,16,251, 4, 9, 16, 25) from the sequence (3,8,15,24,353, 8, 15, 24, 35) to leave 2,4,6,8,102, 4, 6, 8, 10, which has nth term 2n2n.

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

Markers give a mark for the constant second difference, a mark for the n2n^2 coefficient, a mark for the linear remainder 2n2n, and a mark for the full expression n2+2nn^2 + 2n.

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