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Exam questions · Maths · Functions, Sequences and Rates of Change

Geometric and Special Sequences

  • 6 exam questions
  • 16 marks
  • 9 quick checks
  1. 1 Work out [3 marks]

    Here are the first four terms of a geometric sequence: \(2, 6, 18, 54\) (a) Write down the common ratio. (1 mark) (b) Write down the next two terms. (2 marks)

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    Model answer

    (a) \(6 \div 2 = 3\). (b) \(54 \times 3 = 162\) and \(162 \times 3 = 486\).

    Mark scheme

    • (a) \(3\) — B1
    • (b) \(162\) — B1
    • (b) \(486\) — B1
  2. 2 Work out [2 marks]

    The first two terms of a sequence are 3 and 5. Each term after that is the sum of the two terms before it. Work out the 6th term. (2 marks)

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    Model answer

    The terms are \(3, 5, 8, 13, 21, 34\), so the 6th term is 34.

    Mark scheme

    • Continues the sequence, \(8, 13, 21\) — M1
    • \(34\) — A1
  3. 3 Work out [2 marks]

    The first term of a geometric sequence is 5 and the common ratio is 2. Work out the 6th term. (2 marks)

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    Model answer

    \(5 \times 2^5 = 5 \times 32 = 160\).

    Mark scheme

    • \(5 \times 2^5\) or \(5, 10, 20, 40, 80\) — M1
    • \(160\) — A1
  4. 4 Work out [3 marks]

    The 2nd term of a geometric sequence is 6 and the 5th term is 48. Work out the first term. (3 marks)

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    Model answer

    There are 3 steps from the 2nd to the 5th term, so \(r^3 = \dfrac{48}{6} = 8\) and \(r = 2\). The first term is \(6 \div 2 = 3\).

    Mark scheme

    • \(r^3 = \dfrac{48}{6} = 8\) — M1
    • \(r = 2\) — A1
    • \(3\) — A1
  5. 5 Work out [3 marks]

    The first four terms of a geometric sequence are \(2, 2\sqrt{3}, 6, 6\sqrt{3}\). Work out the 7th term. (3 marks)

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    Model answer

    The common ratio is \(\sqrt{3}\). The terms continue \(18, 18\sqrt{3}, 54\), so the 7th term is 54.

    Mark scheme

    • Common ratio \(\sqrt{3}\) — B1
    • \(2 \times (\sqrt{3})^6\) or continues to \(18, 18\sqrt{3}\) — M1
    • \(54\) — A1
  6. 6 Work out [3 marks]

    \(4, x, 36\) are three consecutive terms of a geometric sequence. All the terms are positive. Work out the value of \(x\). (3 marks)

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    Model answer

    \(\dfrac{x}{4} = \dfrac{36}{x}\), so \(x^2 = 144\) and \(x = 12\).

    Mark scheme

    • \(\dfrac{x}{4} = \dfrac{36}{x}\) — M1
    • \(x^2 = 144\) — M1
    • \(12\) — A1

Quick check

  1. 1

    What is the common ratio of \(3, 12, 48, 192\)?

    1. A9
    2. B3
    3. C4
    4. D36
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    C: 4

    \(12 \div 3 = 4\).

  2. 2

    What is the next term of \(2, 6, 18, 54\)?

    1. A108
    2. B162
    3. C72
    4. D216
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    B: 162

    Multiply by 3: \(54 \times 3 = 162\).

  3. 3

    What is the next term in the Fibonacci-type sequence \(3, 5, 8, 13\)?

    1. A21
    2. B18
    3. C26
    4. D16
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    A: 21

    \(8 + 13 = 21\).

  4. 4

    What is the common ratio of \(80, 40, 20, 10\)?

    1. A2
    2. B\(-2\)
    3. C\(\dfrac{1}{4}\)
    4. D\(\dfrac{1}{2}\)
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    D: \(\dfrac{1}{2}\)

    \(40 \div 80 = \dfrac{1}{2}\).

  5. 5

    What is the 5th term of the geometric sequence \(2, 6, 18, \ldots\)?

    1. A54
    2. B486
    3. C162
    4. D90
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    C: 162

    \(2 \times 3^4 = 162\).

  6. 6

    Which of these sequences is geometric?

    1. A\(1, 3, 5, 7\)
    2. B\(1, 3, 9, 27\)
    3. C\(1, 4, 9, 16\)
    4. D\(1, 1, 2, 3\)
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    B: \(1, 3, 9, 27\)

    Each term is multiplied by 3.

  7. 7

    The 2nd term of a geometric sequence is 6 and the 5th term is 48. What is the common ratio?

    1. A2
    2. B3
    3. C8
    4. D4
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    A: 2

    \(r^3 = \dfrac{48}{6} = 8\), so \(r = 2\).

  8. 8

    What is the common ratio of \(2, 2\sqrt{3}, 6, 6\sqrt{3}\)?

    1. A3
    2. B\(2\sqrt{3}\)
    3. C\(\dfrac{1}{\sqrt{3}}\)
    4. D\(\sqrt{3}\)
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    D: \(\sqrt{3}\)

    \(\dfrac{2\sqrt{3}}{2} = \sqrt{3}\).

  9. 9

    A geometric sequence has first term 3 and common ratio 2. What is the \(n\)th term?

    1. A\(3 \times 2^n\)
    2. B\(3n + 2\)
    3. C\(3 \times 2^{n-1}\)
    4. D\(2 \times 3^{n-1}\)
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    C: \(3 \times 2^{n-1}\)

    The \(n\)th term is \(ar^{n-1}\).