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 .
Find the smallest value of n for which Sn>40 .
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]
METHOD 1
setting up an inequality (accept an equation) M1
e.g. 181(3n−1)2>40 , 181(1−3n)−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]
Examiners report
Part (a) was well done.
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.