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, \({u_1} = \frac{1}{{81}}\) and \({u_4} = \frac{1}{3}\) .
Find the value of \(r\) .
Find the smallest value of n for which \({S_n} > 40\) .
Markscheme
evidence of substituting into formula for \(n\)th term of GP (M1)
e.g. \({u_4} = \frac{1}{{81}}{r^3}\)
setting up correct equation \(\frac{1}{{81}}{r^3} = \frac{1}{3}\) A1
\(r = 3\) A1 N2
[3 marks]
METHOD 1
setting up an inequality (accept an equation) M1
e.g. \(\frac{{\frac{1}{{81}}({3^n} - 1)}}{2} > 40\) , \(\frac{{\frac{1}{{81}}(1 - {3^n})}}{{ - 2}} > 40\) , \({3^n} > 6481\)
evidence of solving M1
e.g. graph, taking logs
\(n > 7.9888 \ldots \) (A1)
\(\therefore n = 8\) 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.