| Date | May 2011 | Marks available | 5 | Reference code | 11M.1.hl.TZ1.3 |
| Level | HL only | Paper | 1 | Time zone | TZ1 |
| Command term | Find | Question number | 3 | Adapted from | N/A |
Question
A geometric sequence \({u_1}\) , \({u_2}\) , \({u_3}\) , \(...\) has \({u_1} = 27\) and a sum to infinity of \(\frac{{81}}{2}\).
Find the common ratio of the geometric sequence.
An arithmetic sequence \({v_1}\) , \({v_2}\) , \({v_3}\) , \(...\) is such that \({v_2} = {u_2}\) and \({v_4} = {u_4}\) .
Find the greatest value of \(N\) such that \(\sum\limits_{n = 1}^N {{v_n}} > 0\) .
Markscheme
\({u_1} = 27\)
\(\frac{{81}}{2} = \frac{{27}}{{1 - r}}\) M1
\(r = \frac{1}{3}\) A1
[2 marks]
\({v_2} = 9\)
\({v_4} = 1\)
\(2d = - 8 \Rightarrow d = - 4\) (A1)
\({v_1} = 13\) (A1)
\(\frac{N}{2}\left( {2 \times 13 - 4\left( {N - 1} \right)} \right) > 0\) (accept equality) M1
\(\frac{N}{2}\left( {30 - 4N} \right) > 0\)
\(N\left( {15 - 2N} \right) > 0\)
\(N < 7.5\) (M1)
\(N = 7\) A1
Note: \(13 + 9 + 5 + 1 - 3 - 7 - 11 > 0 \Rightarrow N = 7\) or equivalent receives full marks.
[5 marks]
Examiners report
Part (a) was well done by most candidates. However (b) caused difficulty to most candidates. Although a number of different approaches were seen, just a small number of candidates obtained full marks for this question.
Part (a) was well done by most candidates. However (b) caused difficulty to most candidates. Although a number of different approaches were seen, just a small number of candidates obtained full marks for this question.
