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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]

Find the value of ${S_6}$ .

[2]

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

[3]

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]

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]

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]

Examiners report

[N/A]

Syllabus sections

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

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