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How do you find the nth term of a sequence and recognise special sequences?

Continuing sequences, finding the nth term of linear and quadratic sequences, and recognising geometric, triangular and Fibonacci sequences.

A focused answer to the AQA GCSE Mathematics algebra content on sequences, covering continuing sequences, finding the nth term of linear and quadratic sequences, and recognising geometric, triangular and Fibonacci sequences.

Generated by Claude Opus 4.89 min answer

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  1. What this dot point is asking
  2. Continuing a sequence
  3. The nth term of a linear sequence
  4. Quadratic sequences at Higher tier
  5. Special sequences to recognise
  6. Using the nth term to answer questions
  7. Recognising the type from the differences

What this dot point is asking

AQA wants you to continue a sequence, find a formula for its nnth term, and recognise the named special sequences. Linear (arithmetic) sequences appear at both tiers; quadratic sequences and the geometric, triangular and Fibonacci patterns are Higher-tier favourites. The nnth term turns a list into a rule, letting you find any term without writing out all the ones before it.

Continuing a sequence

To continue a sequence, work out the rule connecting consecutive terms. Most often it is a common difference (add the same amount each time) or a common ratio (multiply by the same amount). For 5,8,11,14,5, 8, 11, 14, \ldots you add 33, so the next term is 1717. For 2,6,18,54,2, 6, 18, 54, \ldots you multiply by 33, so the next term is 162162.

The nth term of a linear sequence

For 7,12,17,22,7, 12, 17, 22, \ldots the common difference is 55, so the nnth term starts as 5n5n. Checking n=1n = 1 gives 55, but the sequence starts at 77, so add 22: the nnth term is 5n+25n + 2. You can then find the 100th term directly as 5(100)+2=5025(100) + 2 = 502, or test whether 158158 is a term by solving 5n+2=1585n + 2 = 158, giving n=31.2n = 31.2, which is not a whole number, so 158158 is not in the sequence.

Quadratic sequences at Higher tier

A quadratic sequence has a constant second difference and an nnth term of the form an2+bn+can^2 + bn + c.

Special sequences to recognise

  • Geometric: a constant ratio between terms, such as 2,6,18,542, 6, 18, 54 (ratio 33).
  • Triangular numbers: 1,3,6,10,15,1, 3, 6, 10, 15, \ldots, formed by adding 1,2,3,1, 2, 3, \ldots; the nnth term is 12n(n+1)\tfrac{1}{2}n(n + 1).
  • Square numbers: 1,4,9,16,1, 4, 9, 16, \ldots with nnth term n2n^2.
  • Cube numbers: 1,8,27,64,1, 8, 27, 64, \ldots with nnth term n3n^3.
  • Fibonacci-type: each term is the sum of the previous two, as in 1,1,2,3,5,8,1, 1, 2, 3, 5, 8, \ldots or any starting pair.

Using the nth term to answer questions

Once you have the nnth term, two standard questions follow. To find a specific term, substitute its position: the 2020th term of 4n+34n + 3 is 4×20+3=834 \times 20 + 3 = 83. To test whether a value is in the sequence, set the nnth term equal to it and solve for nn; if nn is a positive whole number the value is in the sequence, otherwise it is not. For 4n+3=994n + 3 = 99, solving gives n=24n = 24, a whole number, so 9999 is the 2424th term, but 4n+3=1004n + 3 = 100 gives n=24.25n = 24.25, so 100100 is not a term. This "is it a term" test is a reliable exam question.

Recognising the type from the differences

The differences are the diagnostic tool for any sequence. Constant first differences mean a linear sequence and a dn+cdn + c rule. Constant second differences mean a quadratic sequence and an an2+bn+can^2 + bn + c rule. A constant ratio between terms (rather than a constant difference) signals a geometric sequence. Checking the first differences, and then the second differences if the first are not constant, is the systematic first move that tells you which method to apply before committing to any algebra.

Exam-style practice questions

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

AQA 20183 marksA sequence begins 7,11,15,19,7, 11, 15, 19, \ldots Find an expression for the nnth term, and hence find the 50th term. (Foundation tier, Paper 1, non-calculator.)
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The common difference is 44, so the nnth term has the form 4n+c4n + c. Using the first term, 4(1)+c=74(1) + c = 7 gives c=3c = 3, so the nnth term is 4n+34n + 3.

The 50th term is 4(50)+3=2034(50) + 3 = 203.

Markers award a mark for 4n4n, a mark for the correct constant giving 4n+34n + 3, and a mark for 203203. Writing 4n+74n + 7 (using the first term as the constant) is the standard error.

AQA 20214 marksFind the nnth term of the quadratic sequence 3,8,15,24,35,3, 8, 15, 24, 35, \ldots (Higher tier, Paper 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, so the n2n^2 term is n2n^2.

Subtract n2n^2 from each term: 31=23 - 1 = 2, 84=48 - 4 = 4, 159=615 - 9 = 6, 2416=824 - 16 = 8. This linear sequence 2,4,6,82, 4, 6, 8 has nnth term 2n2n.

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

Markers reward identifying the constant second difference, the n2n^2 term, the remaining linear part, and the combined answer.

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