| Date | May 2011 | Marks available | 3 | Reference code | 11M.1.hl.TZ2.10 |
| Level | HL only | Paper | 1 | Time zone | TZ2 |
| Command term | Show that | Question number | 10 | Adapted from | N/A |
Question
An arithmetic sequence has first term a and common difference d, \(d \ne 0\) . The \({{\text{3}}^{{\text{rd}}}}\), \({{\text{4}}^{{\text{th}}}}\) and \({{\text{7}}^{{\text{th}}}}\) terms of the arithmetic sequence are the first three terms of a geometric sequence.
Show that \(a = - \frac{3}{2}d\) .
Show that the \({{\text{4}}^{{\text{th}}}}\) term of the geometric sequence is the \({\text{1}}{{\text{6}}^{{\text{th}}}}\) term of the arithmetic sequence.
Markscheme
let the first three terms of the geometric sequence be given by \({u_1}\) , \({u_1}r\) , \({u_1}{r^2}\)
\(\therefore {u_1} = a + 2d\) , \({u_1}r = a + 3d\) and \({u_1}{r^2} = a + 6d\) (M1)
\(\frac{{a + 6d}}{{a + 3d}} = \frac{{a + 3d}}{{a + 2d}}\) A1
\({a^2} + 8ad + 12{d^2} = {a^2} + 6ad + 9{d^2}\) A1
2a + 3d = 0
\(a = - \frac{3}{2}d\) AG
[3 marks]
\({u_1} = \frac{d}{2}\) , \({u_1}r = \frac{{3d}}{2}\) , \(\left( {{u_1}{r^2} = \frac{{9d}}{2}} \right)\) M1
r = 3 A1
geometric \({{\text{4}}^{{\text{th}}}}\) term \({u_1}{r^3} = \frac{{27d}}{2}\) A1
arithmetic \({\text{1}}{{\text{6}}^{{\text{th}}}}\) term \(a + 15d = - \frac{3}{2}d + 15d\) M1
\( = \frac{{27d}}{2}\) A1
Note: Accept alternative methods.
[3 marks]
Examiners report
This question was done well by many students. Those who did not do it well often became involved in convoluted algebraic processes that complicated matters significantly. There were a number of different approaches taken which were valid.
This question was done well by many students. Those who did not do it well often became involved in convoluted algebraic processes that complicated matters significantly. There were a number of different approaches taken which were valid.
