| 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 \(r\).
Find the value of \({S_6}\).
Find the least value of \(n\) such that \({S_n} > 75\,000\).
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]
