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Date	November 2016	Marks available	3	Reference code	16N.1.sl.TZ0.9
Level	SL only	Paper	1	Time zone	TZ0
Command term	Find	Question number	9	Adapted from	N/A

Question

The first two terms of an infinite geometric sequence, in order, are

$2{\log _2}x,{\text{ }}{\log _2}x$ , where $x > 0$ .

The first three terms of an arithmetic sequence, in order, are

${\log _2}x,{\text{ }}{\log _2}\left( {\frac{x}{2}} \right),{\text{ }}{\log _2}\left( {\frac{x}{4}} \right)$ , where $x > 0$ .

Let ${S_{12}}$ be the sum of the first 12 terms of the arithmetic sequence.

Find $r$ .

[2]

Show that the sum of the infinite sequence is $4{\log _2}x$ .

[2]

Find $d$ , giving your answer as an integer.

[4]

Show that ${S_{12}} = 12{\log _2}x - 66$ .

[2]

is equal to half the sum of the infinite geometric sequence, find $x$ , giving your answer in the form ${2^p}$ , where $p \in \mathbb{Q}$ .

[3]

Markscheme

evidence of dividing terms (in any order) (M1)

eg $\,\,\,\,\,$ $\frac{{{\mu _2}}}{{{\mu _1}}},{\text{ }}\frac{{2{{\log }_2}x}}{{{{\log }_2}x}}$

$r = \frac{1}{2}$ A1 N2

[2 marks]

correct substitution (A1)

eg $\,\,\,\,\,$ $\frac{{2{{\log }_2}x}}{{1 - \frac{1}{2}}}$

correct working A1

eg $\,\,\,\,\,$ $\frac{{2{{\log }_2}x}}{{\frac{1}{2}}}$

${S_\infty } = 4{\log _2}x$ AG N0

[2 marks]

evidence of subtracting two terms (in any order) (M1)

eg $\,\,\,\,\,$ ${u_3} - {u_2},{\text{ }}{\log _2}x - {\log _2}\frac{x}{2}$

correct application of the properties of logs (A1)

eg $\,\,\,\,\,$ ${\log _2}\left( {\frac{{\frac{x}{2}}}{x}} \right),{\text{ }}{\log _2}\left( {\frac{x}{2} \times \frac{1}{x}} \right),{\text{ }}({\log _2}x - {\log _2}2) - {\log _2}x$

correct working (A1)

eg $\,\,\,\,\,$ ${\log _2}\frac{1}{2},{\text{ }} - {\log _2}2$

$d = - 1$ A1 N3

[4 marks]

correct substitution into the formula for the sum of an arithmetic sequence (A1)

eg $\,\,\,\,\,$ $\frac{{12}}{2}\left( {2{{\log }_2}x + (12 - 1)( - 1)} \right)$

correct working A1

eg $\,\,\,\,\,$ $6(2{\log _2}x - 11),{\text{ }}\frac{{12}}{2}(2{\log _2}x - 11)$

$12{\log _2}x - 66$ AG N0

[2 marks]

correct equation (A1)

eg $\,\,\,\,\,$ $12{\log _2}x - 66 = 2{\log _2}x$

correct working (A1)

eg $\,\,\,\,\,$ $10{\log _2}x = 66,{\text{ }}{\log _2}x = 6.6,{\text{ }}{2^{66}} = {x^{10}},{\text{ }}{\log _2}\left( {\frac{{{x^{12}}}}{{{x^2}}}} \right) = 66$

$x = {2^{6.6}}$ (accept $p = \frac{{66}}{{10}}$ ) A1 N2

[3 marks]

Examiners report

[N/A]

Syllabus sections

Topic 1 - Algebra » 1.2 » Laws of exponents; laws of logarithms.

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