Exam questions · Maths · Functions, Sequences and Rates of Change
Geometric and Special Sequences
- 6 exam questions
- 16 marks
- 9 quick checks
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1 Work out [3 marks]
Here are the first four terms of a geometric sequence: \(3, 12, 48, 192\) (a) Write down the common ratio. [1 mark] (b) Work out the next two terms. [2 marks]
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Model answer
(a) \(12 \div 3 = 4\). (b) \(192 \times 4 = 768\) and \(768 \times 4 = 3072\).
Mark scheme
- (a) \(4\) — B1
- (b) \(768\) — B1
- (b) \(3072\) — B1
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2 Work out [2 marks]
The first two terms of a sequence are 2 and 7. 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 \(2, 7, 9, 16, 25, 41\), so the 6th term is 41.
Mark scheme
- Continues the sequence, \(9, 16, 25\) — M1
- \(41\) — A1
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3 Work out [2 marks]
The first term of a geometric sequence is 81 and the common ratio is \(\dfrac{1}{3}\). Work out the 4th term. [2 marks]
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Model answer
\(81, 27, 9, 3\), so the 4th term is 3.
Mark scheme
- \(81 \times \left(\dfrac{1}{3}\right)^3\) or \(81, 27, 9\) — M1
- \(3\) — A1
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4 Work out [3 marks]
The 2nd term of a geometric sequence is 12 and the 5th term is 324. 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 \(r^3 = \dfrac{324}{12} = 27\) and \(r = 3\). The first term is \(12 \div 3 = 4\).
Mark scheme
- \(r^3 = \dfrac{324}{12} = 27\) — M1
- \(r = 3\) — A1
- \(4\) — A1
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5 Work out [3 marks]
The first four terms of a geometric sequence are \(4, 4\sqrt{2}, 8, 8\sqrt{2}\). Work out the 6th term. [3 marks]
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Model answer
The common ratio is \(\sqrt{2}\). The 5th term is \(8\sqrt{2} \times \sqrt{2} = 16\), and the 6th term is \(16\sqrt{2}\).
Mark scheme
- Common ratio \(\sqrt{2}\) — B1
- \(8\sqrt{2} \times \sqrt{2} = 16\) for the 5th term — M1
- \(16\sqrt{2}\) — A1
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6 Work out [3 marks]
\(2, x, 18\) 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
\(\dfrac{x}{2} = \dfrac{18}{x}\), so \(x^2 = 36\) and \(x = 6\).
Mark scheme
- \(\dfrac{x}{2} = \dfrac{18}{x}\) — M1
- \(x^2 = 36\) — M1
- \(6\) — A1
Quick check
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1
What is the common ratio of \(3, 12, 48, 192\)?
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C: 4
\(12 \div 3 = 4\).
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2
What is the next term of \(2, 6, 18, 54\)?
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B: 162
Multiply by 3: \(54 \times 3 = 162\).
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3
What is the next term in the Fibonacci-type sequence \(3, 5, 8, 13\)?
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A: 21
\(8 + 13 = 21\).
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4
What is the common ratio of \(80, 40, 20, 10\)?
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D: \(\dfrac{1}{2}\)
\(40 \div 80 = \dfrac{1}{2}\).
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5
What is the 5th term of the geometric sequence \(2, 6, 18, \ldots\)?
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C: 162
\(2 \times 3^4 = 162\).
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6
Which of these sequences is geometric?
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B: \(1, 3, 9, 27\)
Each term is multiplied by 3.
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7
The 2nd term of a geometric sequence is 6 and the 5th term is 48. What is the common ratio?
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A: 2
\(r^3 = \dfrac{48}{6} = 8\), so \(r = 2\).
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8
What is the common ratio of \(2, 2\sqrt{3}, 6, 6\sqrt{3}\)?
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D: \(\sqrt{3}\)
\(\dfrac{2\sqrt{3}}{2} = \sqrt{3}\).
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9
A geometric sequence has first term 3 and common ratio 2. What is the \(n\)th term?
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C: \(3 \times 2^{n-1}\)
The \(n\)th term is \(ar^{n-1}\).