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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]
a.

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

[2]
b.

Find \(d\), giving your answer as an integer.

[4]
c.

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

[2]
d.

Given that \({S_{12}}\) 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]
e.

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]

a.

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]

b.

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]

c.

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]

d.

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]

e.

Examiners report

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d.
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e.

Syllabus sections

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