Date | May 2017 | Marks available | 4 | Reference code | 17M.1.hl.TZ2.3 |
Level | HL only | Paper | 1 | Time zone | TZ2 |
Command term | Find | Question number | 3 | Adapted from | N/A |
Question
The 1st, 4th and 8th terms of an arithmetic sequence, with common difference \(d\), \(d \ne 0\), are the first three terms of a geometric sequence, with common ratio \(r\). Given that the 1st term of both sequences is 9 find
the value of \(d\);
the value of \(r\);
Markscheme
EITHER
the first three terms of the geometric sequence are \(9\), \(9r\) and \(9{r^2}\) (M1)
\(9 + 3d = 9r( \Rightarrow 3 + d = 3r)\) and \(9 + 7d = 9{r^2}\) (A1)
attempt to solve simultaneously (M1)
\(9 + 7d = 9{\left( {\frac{{3 + d}}{3}} \right)^2}\)
OR
the \({{\text{1}}^{{\text{st}}}}\), \({{\text{4}}^{{\text{th}}}}\) and \({{\text{8}}^{{\text{th}}}}\) terms of the arithmetic sequence are
\(9,{\text{ }}9 + 3d,{\text{ }}9 + 7d\) (M1)
\(\frac{{9 + 7d}}{{9 + 3d}} = \frac{{9 + 3d}}{9}\) (A1)
attempt to solve (M1)
THEN
\(d = 1\) A1
[4 marks]
\(r = \frac{4}{3}\) A1
Note: Accept answers where a candidate obtains \(d\) by finding \(r\) first. The first two marks in either method for part (a) are awarded for the same ideas and the third mark is awarded for attempting to solve an equation in \(r\).
[1 mark]