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Exam questions · Maths · Algebra

Sequences and the nth Term

  • 7 exam questions
  • 23 marks
  • 10 quick checks
  1. 1 Find [2 marks]

    Find an expression for the nth term of the sequence \(5, 12, 19, 26, \ldots\)

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    Model answer

    The difference is 7, so the nth term begins \(7n\). \(7n\) is \(7, 14, 21, 28\), and each term is 2 less, so the nth term is \(7n - 2\).

    Mark scheme

    • \(7n\) seen — M1
    • \(7n - 2\) — A1
  2. 2 Work out [5 marks]

    Here are the first three patterns in a sequence. The patterns are made from matchsticks. (a) Work out the number of matchsticks in Pattern 10. [1 mark] (b) Find an expression, in terms of \(n\), for the number of matchsticks in Pattern \(n\). [2 marks] (c) Which pattern number uses exactly 70 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

    (b) Each new square needs 3 more matchsticks and Pattern 1 has 4, so the nth term is \(3n + 1\). (a) \(3 \times 10 + 1 = 31\). (c) \(3n + 1 = 70\), so \(3n = 69\) and \(n = 23\). It is Pattern 23.

    Mark scheme

    • (a) 31, or \(3 \times 10 + 1\) — B1
    • (b) \(3n\) seen — M1
    • (b) \(3n + 1\) — A1
    • (c) \(3n + 1 = 70\) or \(3n = 69\) — M1
    • (c) Pattern 23 — A1
  3. 3 Work out [4 marks]

    The nth term of a sequence is \(2n^2 + 1\). (a) Work out the first three terms of the sequence. [2 marks] (b) Is 51 a term of this sequence? You must show how you decide. [2 marks]

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    Model answer

    (a) \(2 \times 1^2 + 1 = 3\), \(2 \times 2^2 + 1 = 9\) and \(2 \times 3^2 + 1 = 19\). (b) \(2n^2 + 1 = 51\) gives \(2n^2 = 50\), so \(n^2 = 25\) and \(n = 5\). Yes, 51 is the 5th term.

    Mark scheme

    • (a) Two terms correct — M1
    • (a) 3, 9, 19 — A1
    • (b) \(2n^2 = 50\) or \(n^2 = 25\) — M1
    • (b) Yes, it is the 5th term — A1
  4. 4 Work out [3 marks]

    Here are the first four terms of a geometric sequence. \(2\) \(6\) \(18\) \(54\) (a) Write down the next term. [1 mark] (b) Work out the 7th term. [2 marks]

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    Model answer

    (a) Each term is multiplied by 3, so the next term is \(54 \times 3 = 162\). (b) The 7th term is \(2 \times 3^6 = 2 \times 729 = 1458\).

    Mark scheme

    • (a) 162 — B1
    • (b) \(2 \times 3^6\) or continues the sequence to 486 and 1458 — M1
    • (b) 1458 — A1
  5. 5 Work out [2 marks]

    The nth term of a sequence is \(3n + 5\). Sally says that 80 is a term in this sequence. Is Sally correct? You must show how you decide.

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    Model answer

    Solve \(3n + 5 = 80\): \(3n = 75\), so \(n = 25\). Since 25 is a whole number, 80 is the 25th term, so Sally is correct.

    Mark scheme

    • \(3n + 5 = 80\) or \(3n = 75\) — M1
    • Yes, with \(n = 25\) — A1
  6. 6 Find [3 marks]

    Here are the first four terms of a quadratic sequence. \(3\) \(10\) \(21\) \(36\) Find an expression for the nth term.

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    Model answer

    The first differences are 7, 11, 15 and the second difference is 4, so the nth term starts \(2n^2\), which is 2, 8, 18, 32. Subtracting leaves 1, 2, 3, 4, which is \(n\). The nth term is \(2n^2 + n\).

    Mark scheme

    • Second difference 4 found, or \(2n^2\) seen — M1
    • Subtracts \(2n^2\) to leave 1, 2, 3, 4 — M1
    • \(2n^2 + n\) — A1
  7. 7 Work out [4 marks]

    Here are the first three terms of a sequence. \(17\) \(13\) \(9\) (a) Find an expression for the nth term. [2 marks] (b) Work out the first term in the sequence that is negative. [2 marks]

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    Model answer

    (a) The sequence goes down by 4, so it begins \(-4n\). \(-4n\) is \(-4, -8, -12\), and each term is 21 more, so the nth term is \(-4n + 21\). (b) \(-4n + 21 < 0\) gives \(n > 5.25\), so the first negative term is when \(n = 6\): \(-4 \times 6 + 21 = -3\).

    Mark scheme

    • (a) \(-4n\) seen — M1
    • (a) \(-4n + 21\) — A1
    • (b) \(-4n + 21 < 0\), or finds the terms 5 and 1 then \(-3\) — M1
    • (b) \(-3\) — 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 \(n^2\).