Date | May 2016 | Marks available | 6 | Reference code | 16M.2.hl.TZ2.4 |
Level | HL only | Paper | 2 | Time zone | TZ2 |
Command term | Find | Question number | 4 | Adapted from | N/A |
Question
The sum of the second and third terms of a geometric sequence is 96.
The sum to infinity of this sequence is 500.
Find the possible values for the common ratio, \(r\).
Markscheme
\(ar + a{r^2} = 96\) A1
Note: Award A1 for any valid equation involving \(a\) and \(r\), eg, \(\frac{{a(1 - {r^3})}}{{1 - r}} - a = 96\).
\(\frac{a}{{1 - r}} = 500\) A1
EITHER
attempting to eliminate \(a\) to obtain \(500r(1 - {r^2}) = 96\) (or equivalent in unsimplified form) (M1)
OR
attempting to obtain \(a = \frac{{96}}{{r + {r^2}}}\) and \(a = 500(1 - r)\) (M1)
THEN
attempting to solve for \(r\) (M1)
\(r = 0.2{\text{ }}\left( { = \frac{1}{5}} \right)\) or \(r = 0.885{\text{ }}\left( { = \frac{{\sqrt {97} - 1}}{{10}}} \right)\) A1A1
[6 marks]
Examiners report
Reasonably well done. Quite a number of candidates included a solution outside \( - 1 < r < 1\).