C2 January 2009 Q9
9. The first three terms of a geometric series are \((k + 4)\), \(k\) and \((2k - 15)\) respectively, where \(k\) is a positive constant.
| Scheme | Marks |
|---|---|
| Initial step: Two of: \(a = k + 4,\ \ ar = k,\ \ ar^2 = 2k - 15\) Or one of: \(r = \dfrac{k}{k + 4},\quad r = \dfrac{2k - 15}{k},\quad r^2 = \dfrac{2k - 15}{k + 4}\) , Or \(k = \sqrt{(k + 4)(2k - 15)}\) or even \(k^3 = (k + 4)k(2k - 15)\) | M1 |
| \(k^2 = (k + 4)(2k - 15)\), so \(k^2 = 2k^2 + 8k - 15k - 60\) | M1, A1 |
| Proceed to \(k^2 - 7k - 60 = 0\) (*) | A1 |
| (4) |
Notes
M1: The ‘initial step’, scoring the first M mark, may be implied by next line of proof
M1: Eliminates \(a\) and \(r\) to give valid equation in \(k\) only. Can be awarded for equation involving fractions.
A1 : need some correct expansion and working and answer equivalent to required quadratic but with uncollected terms. Equations involving fractions do not get this mark. (No fractions, no brackets – could be a cubic equation)
A1: as answer is printed this mark is for cso (Needs = 0)
All four marks must be scored in part (a)
| Scheme | Marks |
|---|---|
| \((k - 12)(k + 5) = 0\qquad k = 12\) (*) | M1 A1 |
| (2) |
Notes
M1: Attempt to solve quadratic
A1: This is for correct factorisation or solution and \(k = 12\). Ignore the extra solution (\(k = -5\) or even \(k = 5\)), if seen.
Substitute and verify is M1 A0
Marks must be scored in part (b)
| Scheme | Marks |
|---|---|
| Common ratio: \(\dfrac{k}{k + 4}\) or \(\dfrac{2k - 15}{k} = \dfrac{12}{16}\ \left(= \dfrac{3}{4}\ \text{or}\ 0.75\right)\) | M1 A1 |
| (2) |
Notes
M1: Complete method to find \(r\) Could have answer in terms of \(k\)
A1: 0.75 or any correct equivalent
Both Marks must be scored in (c)
| Scheme | Marks |
|---|---|
| \(\dfrac{a}{1 - r} = \dfrac{16}{\left(\tfrac{1}{4}\right)} = 64\) | M1 A1 |
| (2) | |
| [10] |
Notes
M1: Tries to use \(\dfrac{a}{1 - r}\), (even with \(r > 1\)). Could have an answer still in terms of \(k\).
A1: This answer is 64 cao.