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Date November 2015 Marks available 3 Reference code 15N.2.sl.TZ0.4
Level SL only Paper 2 Time zone TZ0
Command term Find Question number 4 Adapted from N/A

Question

The first three terms of a geometric sequence are \({u_1} = 0.64,{\text{ }}{u_2} = 1.6\), and \({u_3} = 4\).

Find the value of \(r\).

[2]
a.

Find the value of \({S_6}\).

[2]
b.

Find the least value of \(n\) such that \({S_n} > 75\,000\).

[3]
c.

Markscheme

valid approach     (M1)

eg\(\;\;\;\frac{{{u_1}}}{{{u_2}}},{\text{ }}\frac{4}{{1.6}},{\text{ }}1.6 = r(0.64)\)

\(r = 2.5\;\;\;\left( { = \frac{5}{2}} \right)\)     A1     N2

[2 marks]

a.

correct substitution into \({S_6}\)     (A1)

eg\(\;\;\;\frac{{0.64({{2.5}^6} - 1)}}{{2.5 - 1}}\)

\({S_6} = 103.74\) (exact), \(104\)     A1     N2

[2 marks]

b.

METHOD 1 (analytic)

valid approach     (M1)

eg\(\;\;\;\frac{{0.64({{2.5}^n} - 1)}}{{2.5 - 1}} > 75\,000,{\text{ }}\frac{{0.64({{2.5}^n} - 1)}}{{2.5 - 1}} = 75\,000\)

correct inequality (accept equation)     (A1)

eg\(\;\;\;n > 13.1803,{\text{ }}n = 13.2\)

\(n = 14\)     A1     N1

METHOD 2 (table of values)

both crossover values     A2

eg\(\;\;\;{S_{13}} = 63577.8,{\text{ }}{S_{14}} = 158945\)

\(n = 14\)     A1     N1

[3 marks]

Total [7 marks]

c.

Examiners report

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a.
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b.
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c.

Syllabus sections

Topic 2 - Functions and equations » 2.7 » Solving equations, both graphically and analytically.
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