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Date May 2009 Marks available 4 Reference code 09M.2.sl.TZ1.6
Level SL only Paper 2 Time zone TZ1
Command term Find Question number 6 Adapted from N/A

Question

In a geometric series, u1=181 and u4=13 .

Find the value of r .

[3]
a.

Find the smallest value of n for which Sn>40 .

[4]
b.

Markscheme

evidence of substituting into formula for nth term of GP     (M1)

e.g. u4=181r3

setting up correct equation 181r3=13     A1

r=3     A1     N2

[3 marks]

a.

METHOD 1

setting up an inequality (accept an equation)     M1

e.g. 181(3n1)2>40 , 181(13n)2>40 , 3n>6481

evidence of solving     M1

e.g. graph, taking logs

n>7.9888     (A1)

     A1     N2

METHOD 2

if n = 7 , sum = 13.49 \ldots  ; if n = 8 , sum = 40.49 \ldots      A2

n = 8 (is the smallest value)     A2     N2

[4 marks]

b.

Examiners report

Part (a) was well done.

a.

In part (b) a good number of candidates did not realize that they could use logs to solve the problem, nor did they make good use of their GDCs. Some students did use a trial and error approach to check various values however, in many cases, they only checked one of the "crossover" values. Most candidates had difficulty with notation, opting to set up an equation rather than an inequality.

b.

Syllabus sections

Topic 1 - Algebra » 1.1 » Arithmetic sequences and series; sum of finite arithmetic series; geometric sequences and series; sum of finite and infinite geometric series.
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