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Exam questions · Maths · Algebra

Sequences and the nth Term

  • 7 exam questions
  • 21 marks
  • 10 quick checks
  1. 1 Find [3 marks]

    Here are the first five terms of a number sequence. \(4\) \(9\) \(14\) \(19\) \(24\) (a) Write down the next term of the sequence. (1 mark) (b) Find an expression, in terms of \(n\), for the nth term of the sequence. (2 marks)

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    Model answer

    (a) The sequence goes up by 5, so the next term is \(24 + 5 = 29\). (b) The difference is 5, so the nth term starts \(5n\). The sequence \(5n\) is 5, 10, 15, 20, and the terms are each 1 less, so the nth term is \(5n - 1\).

    Mark scheme

    • (a) 29 — B1
    • (b) \(5n\) seen — M1
    • (b) \(5n - 1\) — A1
  2. 2 Work out [5 marks]

    Here are the first three patterns in a sequence. Each pattern is made using matchsticks. (a) Work out the number of matchsticks in Pattern 4. (1 mark) (b) Find an expression, in terms of \(n\), for the number of matchsticks in Pattern \(n\). (2 marks) One of the patterns uses exactly 49 matchsticks. (c) Work out which pattern uses exactly 49 matchsticks. (2 marks)

    The first three patterns in a sequence: one square, two squares and three squares in a row, made from matchsticks.
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    Model answer

    (a) The numbers of matchsticks are 4, 7, 10, so the next is 13. (b) Each new square adds 3 matchsticks, so the nth term starts \(3n\), and \(3n + 1\) fits: \(3 \times 1 + 1 = 4\). (c) \(3n + 1 = 49\), so \(3n = 48\) and \(n = 16\). It is Pattern 16.

    Mark scheme

    • (a) 13 — B1
    • (b) \(3n\) seen — M1
    • (b) \(3n + 1\) — A1
    • (c) \(3n + 1 = 49\) or \(3n = 48\) — M1
    • (c) Pattern 16 — A1
  3. 3 Find [2 marks]

    Find an expression, in terms of \(n\), for the nth term of the sequence \(11, 8, 5, 2, \ldots\)

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    Model answer

    The sequence goes down by 3 each time, so the nth term begins \(-3n\). \(-3n\) is \(-3, -6, -9, -12\), and the terms are each 14 more, so the nth term is \(-3n + 14\).

    Mark scheme

    • \(-3n\) seen — M1
    • \(-3n + 14\) or \(14 - 3n\) — A1
  4. 4 Work out [3 marks]

    The nth term of a sequence is \(4n + 3\). (a) Work out the 25th term. (1 mark) (b) Is 150 a term of this sequence? You must show how you decide. (2 marks)

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    Model answer

    (a) \(4 \times 25 + 3 = 103\). (b) Solve \(4n + 3 = 150\): \(4n = 147\), so \(n = 36.75\). Since \(n\) must be a whole number, 150 is not a term of the sequence.

    Mark scheme

    • (a) 103 — B1
    • (b) \(4n + 3 = 150\) or \(4n = 147\) — M1
    • (b) No, with a reason such as n = 36.75 not being an integer — C1
  5. 5 Write down [2 marks]

    Here are the first five terms of a Fibonacci-type sequence. \(2\) \(3\) \(5\) \(8\) \(13\) Write down the next two terms of the sequence and describe the rule you used.

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    Model answer

    The next two terms are 21 and 34. Each term is the sum of the two terms before it: \(8 + 13 = 21\) and \(13 + 21 = 34\).

    Mark scheme

    • A correct description of the rule: add the previous two terms — M1
    • 21 and 34 — A1
  6. 6 Find [3 marks]

    Here are the first four terms of a quadratic sequence. \(4\) \(10\) \(18\) \(28\) Find an expression for the nth term of the sequence.

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    Model answer

    The first differences are 6, 8, 10 and the second difference is 2, so the nth term starts \(n^2\). Subtracting \(n^2\) (1, 4, 9, 16) leaves 3, 6, 9, 12, which is \(3n\). The nth term is \(n^2 + 3n\).

    Mark scheme

    • Finds the first or second differences, such as 2 as the second difference — M1
    • \(n^2\) seen and subtracted, leaving \(3n\) — M1
    • \(n^2 + 3n\) — A1
  7. 7 Work out [3 marks]

    Here are the first four terms of a geometric sequence. \(3\) \(6\) \(12\) \(24\) (a) Write down the next two terms. (1 mark) (b) Work out the 8th term. (2 marks)

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    Model answer

    (a) Each term is multiplied by 2, so the next two terms are 48 and 96. (b) The 8th term is \(3 \times 2^7 = 3 \times 128 = 384\).

    Mark scheme

    • (a) 48 and 96 — B1
    • (b) \(3 \times 2^7\) or continues the sequence to the 7th term — M1
    • (b) 384 — A1

Quick check

  1. 1

    The nth term of a sequence is \(3n + 2\). What is the first term that is greater than 100?

    1. A104
    2. B98
    3. C101
    4. D100
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    C: 101

    \(3n + 2 > 100\) gives \(n > 32.67\ldots\), so \(n = 33\) and the term is \(3 \times 33 + 2 = 101\).

  2. 2

    What is the common ratio of the geometric sequence 5, 10, 20, 40, ...?

    1. A10
    2. B2
    3. C5
    4. D15
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    B: 2

    Each term is the previous term multiplied by 2.

  3. 3

    What is the next term in the sequence 2, 5, 8, 11, ...?

    1. A15
    2. B17
    3. C13
    4. D14
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    D: 14

    The sequence goes up by 3 each time, so the next term is \(11 + 3 = 14\).

  4. 4

    What is the nth term of the sequence 4, 7, 10, 13, ...?

    1. A\(4n + 3\)
    2. B\(3n + 4\)
    3. C\(3n + 1\)
    4. D\(n + 3\)
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    C: \(3n + 1\)

    The difference is 3, so it begins \(3n\). The sequence is 1 more than \(3n\), so the nth term is \(3n + 1\).

  5. 5

    Which sequence has nth term \(2n - 3\)?

    1. A\(-1, 1, 3, 5\)
    2. B\(2, 4, 6, 8\)
    3. C\(5, 7, 9, 11\)
    4. D\(-3, -1, 1, 3\)
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    A: \(-1, 1, 3, 5\)

    Substituting \(n = 1, 2, 3, 4\) gives \(-1, 1, 3, 5\).

  6. 6

    What is the nth term of the sequence 20, 17, 14, 11, ...?

    1. A\(-3n + 23\)
    2. B\(-3n + 20\)
    3. C\(23n - 3\)
    4. D\(3n + 17\)
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    A: \(-3n + 23\)

    The sequence goes down by 3, so it begins \(-3n\). Adding 23 gives 20 when \(n = 1\).

  7. 7

    Which of these numbers is a term in the sequence with nth term \(4n + 1\)?

    1. A63
    2. B61
    3. C62
    4. D64
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    B: 61

    \(4n + 1 = 61\) gives \(n = 15\). The other numbers do not give a whole number for \(n\).

  8. 8

    Which of these is a geometric sequence?

    1. A\(3, 6, 9, 12\)
    2. B\(3, 6, 12, 24\)
    3. C\(1, 4, 9, 16\)
    4. D\(1, 1, 2, 3\)
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    B: \(3, 6, 12, 24\)

    Each term is multiplied by 2: \(3 \times 2 = 6\), \(6 \times 2 = 12\), \(12 \times 2 = 24\).

  9. 9

    The first four terms of a Fibonacci-type sequence are 1, 3, 4, 7. What is the 5th term?

    1. A11
    2. B12
    3. C14
    4. D10
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    A: 11

    Each term is the sum of the two before it, so the 5th term is \(4 + 7 = 11\).

  10. 10

    A sequence has a constant second difference of 2. Its nth term begins with which term?

    1. A\(4n^2\)
    2. B\(n^2\)
    3. C\(2n\)
    4. D\(2n^2\)
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    B: \(n^2\)

    The coefficient of \(n^2\) is half the second difference, so it is \(1\), giving \(n^2\).