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How do you continue a sequence and find the nth term of linear and quadratic sequences?

Continue sequences using a term-to-term rule, find and use the nth term of a linear sequence, recognise quadratic and special sequences, and find the nth term of a quadratic sequence (Higher tier).

A CCEA GCSE Mathematics answer on sequences, covering term-to-term rules, finding and using the nth term of a linear sequence, recognising special sequences, and finding the nth term of a quadratic sequence using second differences.

Generated by Claude Opus 4.810 min answer

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  1. What this dot point is asking
  2. Term-to-term and position-to-term rules
  3. The nth term of a linear sequence
  4. Special sequences
  5. Quadratic sequences (Higher)
  6. Why this matters

What this dot point is asking

Sequences are ordered lists of numbers built by a rule, and the CCEA Algebra strand asks you to continue them and to capture the rule algebraically. You must continue a sequence using a term-to-term rule, find and use the nth term (position-to-term rule) of a linear sequence, recognise special sequences such as square, cube and triangular numbers, and at Higher tier find the nth term of a quadratic sequence using second differences. The nth term lets you jump to any term without listing them all, which is the key exam skill.

Term-to-term and position-to-term rules

A term-to-term rule tells you how each term follows from the previous one, such as "add 3" or "multiply by 2". It is easy to continue a sequence with, but slow if you want a distant term.

A position-to-term rule, the nth term, gives the value at any position nn directly. For the sequence 4,7,10,13,4, 7, 10, 13, \ldots the nth term is 3n+13n + 1, so the 100th term is 3×100+1=3013 \times 100 + 1 = 301 without listing the terms in between. This is why the nth term is the powerful tool.

The nth term of a linear sequence

A linear (arithmetic) sequence goes up or down by the same amount each time, the common difference. The nth term always begins with the common difference times nn.

So for 9,14,19,24,9, 14, 19, 24, \ldots the difference is 55, 5n5n gives 5,10,15,205, 10, 15, 20, and each term is 44 more, so the nth term is 5n+45n + 4. A negative common difference works the same way: for 20,17,14,20, 17, 14, \ldots the nth term is 3n+23-3n + 23.

Special sequences

Some sequences are worth recognising on sight. The square numbers are 1,4,9,16,1, 4, 9, 16, \ldots with nth term n2n^2; the cube numbers are 1,8,27,64,1, 8, 27, 64, \ldots with 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 two previous terms. Spotting these speeds up many questions.

Quadratic sequences (Higher)

A quadratic sequence does not have a constant first difference, but its second differences are constant.

Why this matters

Sequences train you to spot pattern and structure and to express it algebraically, which is exactly the reasoning CCEA wants. The linear nth term connects to straight-line graphs (the common difference is the gradient), and quadratic sequences connect to quadratic graphs and expansion. Recognising the special number sequences also pays off in number and geometry questions.

Exam-style practice questions

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

CCEA 20203 marksFind the nth term of the sequence 5,8,11,14,5, 8, 11, 14, \ldots and use it to find the 50th term. (Non-calculator.)
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The common difference is +3+3, so the nth term starts with 3n3n.

Compare 3n3n (3,6,9,3, 6, 9, \ldots) with the sequence (5,8,11,5, 8, 11, \ldots): each term is 2 more, so the nth term is 3n+23n + 2.

The 50th term is 3×50+2=1523 \times 50 + 2 = 152.

Marks are for 3n3n, for the full rule 3n+23n + 2, and for 152152. A common error is to write 3n+53n + 5, forgetting to compare with 3n3n rather than the first term alone.

CCEA 20224 marksFind the nth term of the quadratic sequence 3,8,15,24,3, 8, 15, 24, \ldots. (Higher, non-calculator.)
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The first differences are 5,7,95, 7, 9 and the second differences are constant at 22.

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

Subtract n2n^2 (1,4,9,161, 4, 9, 16) from the sequence (3,8,15,243, 8, 15, 24): the remainders are 2,4,6,82, 4, 6, 8, which is the linear part 2n2n.

So the nth term is n2+2nn^2 + 2n. Marks are for the second difference, the n2n^2 coefficient, the linear part, and the final rule. Checking n=4n = 4 gives 16+8=2416 + 8 = 24, correct.

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