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Date	May 2017	Marks available	2	Reference code	17M.1.SL.TZ1.T_10
Level	Standard Level	Paper	Paper 1 (with calculator from previous syllabus)	Time zone	Time zone 1
Command term	Find	Question number	T_10	Adapted from	N/A

Question

The first three terms of a geometric sequence are ${u_1} = 486,{\text{ }}{u_2} = 162,{\text{ }}{u_3} = 54$ .

Find the value of $r$ , the common ratio of the sequence.

[2]

Find the value of $n$ for which ${u_n} = 2$ .

[2]

Find the sum of the first 30 terms of the sequence.

[2]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$\frac{{162}}{{486}}$ $\,\,\,$ OR $\,\,\,$ $\frac{{54}}{{162}}$ (M1)

Note: Award (M1) for dividing any ${u_{n + 1}}$ by ${u_n}$ .

$= \frac{1}{3}{\text{ }}(0.333,{\text{ }}0.333333 \ldots )$ (A1) (C2)

[2 marks]

$486{\left( {\frac{1}{3}} \right)^{n - 1}} = 2$ (M1)

Note: Award (M1) for their correct substitution into geometric sequence formula.

$n = 6$ (A1)(ft) (C2)

Note: Follow through from part (a).

Award (A1)(A0) for ${u_6} = 2$ or ${u_6}$ with or without working.

[2 marks]

${S_{30}} = \frac{{486\left( {1 - {{\frac{1}{3}}^{30}}} \right)}}{{1 - \frac{1}{3}}}$ (M1)

Note: Award (M1) for correct substitution into geometric series formula.

$= 729$ (A1)(ft) (C2)

[2 marks]

Examiners report

[N/A]

Syllabus sections

Topic 1—Number and algebra » SL 1.3—Geometric sequences and series

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