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A23–A25Sequences and the nth term

Edexcel GCSE Maths Foundation (1MA1), Foundation tier · Algebra

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Revision notes

Generating terms from term-to-term and position-to-term rules, recognising special sequences, and finding the nth term of linear sequences (both tiers) and quadratic sequences (Higher). Includes arithmetic, geometric and Fibonacci-type sequences.

Grade by grade

What you need to be able to do, from the first marks up to the top grade.

  1. 2
    Continue a sequence using a term-to-term rulee.g. 'add 4' starting at 3 gives 3, 7, 11, 15, …; 'multiply by 2' starting at 5 gives 5, 10, 20, 40, …
  2. 3
    Recognise square, cube and triangular numbersSquares 1, 4, 9, 16, …; cubes 1, 8, 27, 64, …; triangular numbers 1, 3, 6, 10, …
  3. 4
    Generate terms from an nth termSubstitute \(n = 1, 2, 3, \ldots\), e.g. \(3n + 2\) gives 5, 8, 11, …
  4. 4
    Find the nth term of a linear sequenceThe common difference multiplies n, then adjust: 5, 9, 13, 17, … has nth term \(4n + 1\).
  5. 5
    Decide if a number is in a sequenceSet the nth term equal to the number and check whether n is a positive whole number.
  6. 5
    Continue Fibonacci-type and geometric sequencesFibonacci-type: add the previous two terms. Geometric: multiply by the same number each time.

Notes

Types of sequence

  • Arithmetic (linear): add or subtract the same amount each time, e.g. 7, 10, 13, 16 (common difference 3).
  • Geometric: multiply by the same number r each time, e.g. 3, 6, 12, 24 (r = 2) or 80, 40, 20, 10 (\(r = \frac{1}{2}\)).
  • Fibonacci-type: each term is the sum of the two before it, e.g. 1, 1, 2, 3, 5, 8.
  • Quadratic: the differences change but the second differences are constant, e.g. 2, 5, 10, 17 (differences 3, 5, 7).
  • Special sequences: square numbers 1, 4, 9, 16, 25; cube numbers 1, 8, 27, 64; triangular numbers 1, 3, 6, 10, 15.

Linear sequences and the nth term

  • A term-to-term rule links each term to the one before. The nth term (position-to-term rule) gives any term from its position n.
  • Find the common difference d. The nth term is \(dn + c\), where c = first term − d.
  • 5, 8, 11, 14: \(d = 3\) and \(c = 5 - 3 = 2\), so the nth term is \(3n + 2\).
  • Decreasing sequences have a negative difference: 20, 16, 12, 8 has nth term \(24 - 4n\).
  • Is 100 in the sequence \(3n + 2\)? \(3n + 2 = 100\) gives \(n = \frac{98}{3}\), which is not a whole number, so no.

Cheatsheet

  • Linear nth term: \(dn + c\), where d is the common difference and c = first term − d
  • Square numbers \(n^2\), cube numbers \(n^3\), triangular numbers \(\frac{n(n + 1)}{2}\)
  • Geometric: multiply by the same number each time, e.g. 2, 4, 8, 16 has nth term \(2^n\)
  • Fibonacci-type: add the previous two terms
  • Is x a term? Solve nth term = x; n must be a positive whole number

How to answer each type of question

Find the nth term (linear)

2 marks4
  1. Find the common difference: this is the number in front of n.
  2. Compare with the times table to find what to add or subtract.
  3. Check with the first two terms.

Example. Here are the first five terms of a sequence: 2, 9, 16, 23, 30. Find an expression, in terms of n, for the nth term.

Show the model answer
Common difference 7, so \(7n + k\) M1
\(7n - 5\) A1

Is this number a term? Give a reason

2 marks5
  1. Set the nth term equal to the number.
  2. Solve for n.
  3. Conclude: it is a term only if n is a positive whole number.

Example. The nth term of a sequence is \(4n + 3\). Is 125 a term in the sequence? Give a reason for your answer.

Show the model answer
\(4n + 3 = 125\) gives \(n = 30.5\) M1
No, because n is not a whole number (the 30th term is 123 and the 31st is 127) C1

Fibonacci-type sequence with an unknown

3 marks5
  1. Write the terms in terms of the unknown, adding the two before each time.
  2. Set the given term equal to its value.
  3. Solve.

Example. The rule for a sequence is: add the two previous terms. The first term is 3 and the second term is b. The fifth term is 21. Find b.

Show the model answer
Terms: 3, \(b\), \(3 + b\), \(3 + 2b\), \(6 + 3b\) M1
\(6 + 3b = 21\) M1
\(b = 5\) A1

Shortcuts and memory tricks

  • Check an nth term by putting in \(n = 1\) and \(n = 2\): you should get the first two terms.
  • Zero term trick: work back one step from the first term to find c. For 5, 8, 11, the zero term is 2, so the nth term is \(3n + 2\).
  • \(3n + 2\) is the 3 times table with 2 added to each term.
  • Higher: 'half the second difference' gives the \(n^2\) coefficient.

Where marks are lost

  • Writing \(n + 3\) for a sequence that goes up in 3s: the difference multiplies n, so it is \(3n + \ldots\)
  • Giving the term-to-term rule ('add 3') when the question asks for the nth term.
  • Forgetting the negative for decreasing sequences: 20, 16, 12 has nth term \(24 - 4n\), not \(4n + 24\).
  • Higher: using the whole second difference as the \(n^2\) coefficient instead of half of it.

Exam technique

  • 'Find an expression in terms of n' needs an answer containing n, e.g. \(6n - 1\), not 'add 6'.
  • When asked 'is x a term? Give a reason', show your calculation and finish with a clear yes or no.
  • Read whether you are given terms or the nth term before you start.

Sample questions

Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.

Question 1Easy3 marks
(a) The \(n\)th term of a sequence is \(2n + 9\).
Write down the first three terms of the sequence.[2]
(b) Work out the 13th term of the sequence.[1]
Show the answer and mark scheme
(a) Answer: \(11,\quad 13,\quad 15\)
  • B2 for \(11, 13, 15\) (B1 for two correct terms)

Worked solution: \(n = 1: \ 2 \times 1 + 9 = 11\)
\(n = 2: \ 2 \times 2 + 9 = 13\)
\(n = 3: \ 2 \times 3 + 9 = 15\)

(b) Answer: \(35\)
  • B1 for 35

Worked solution: \(2 \times 13 + 9 = 35\)

Question 2Medium4 marks
Here are the first five terms of an arithmetic sequence.
\(4,\quad 6,\quad 8,\quad 10,\quad 12\)
(a) Find an expression, in terms of \(n\), for the \(n\)th term of the sequence.[2]
(b) Is \(129\) a term of this sequence?
You must show how you get your answer.[2]
Show the answer and mark scheme
(a) Answer: \(2n + 2\)
  • M1 for \(2n + k\) for any value of \(k\), or a common difference of \(2\) used
  • A1 for \(2n + 2\) oe

Worked solution: The common difference is \(2\), so the \(n\)th term is \(2n + k\).
When \(n = 1\): \(2 + k = 4\) so \(k = 2\). \(n\)th term \(= 2n + 2\).

(b) Answer: No: \(n = 63.5\) is not a whole number, so 129 is not a term.
  • M1 for \(2n + 2 = 129\) or a correct method to find the terms either side of 129
  • C1 for no, with a correct reason, e.g. \(n = 63.5\) is not an integer, or the 63rd and 64th terms are 128 and 130

Worked solution: \(2n + 2 = 129\) gives \(2n = 127\) so \(n = 63.5\).
No: \(n = 63.5\) is not a whole number, so 129 is not a term.

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