C2 June 2009 Q5

Notes

M1   for forming an equation for \(r^3\) based on 96 and 324 (e.g. \(96r^3 = 324\) scores M1).
The equation must involve multiplication/division rather than addition/subtraction.
A1   Do not penalise solutions with working in decimals, providing these are correctly rounded or truncated to at least 2dp and the final answer 2/3 is seen.

Alternative: (verification)
M1   Using \(r^3 = \dfrac{8}{27}\) and multiplying 324 by this (or multiplying by \(r = \dfrac{2}{3}\) three times).
A1   Obtaining 96 (cso). (A conclusion is not required).
\(324 \times \left(\dfrac{2}{3}\right)^3 = 96\) (no real evidence of calculation) is not quite enough and scores M1 A0.

Notes

M1   for the use of a correct formula or for 'working back' by dividing by \(\dfrac{2}{3}\) (or by their \(r\)) twice from 324 (or 5 times from 96).
Exceptionally, allow M1 also for using \(ar^3 = 324\) or \(ar^6 = 96\) instead of \(ar^2 = 324\) or \(ar^5 = 96\), or for dividing by \(r\) three times from 324 (or 6 times from 96)… but no other exceptions are allowed.

Notes

M1   for use of sum to 15 terms formula with values of \(a\) and \(r\). If the wrong power is used, e.g. 14, the M mark is scored only if the correct sum formula is stated.
1^st A1ft for a correct expression or correct ft their \(a\) with \(r = \dfrac{2}{3}\).
2^nd A1   for awrt 2180, even following 'minor inaccuracies'.
Condone missing brackets round the \(\dfrac{2}{3}\) for the marks in part (c).

Alternative:
M1   for adding 15 terms and 1^st A1ft for adding the 15 terms that ft from their \(a\) and \(r = \dfrac{2}{3}\).

(corrected from the printed mark scheme: the marks column prints only “M1A1ft, (3)” for part (c); the final A1 for awrt 2180, described in the notes, has been added.)

Notes

M1   for use of correct sum to infinity formula with their \(a\). For this mark, if a value of \(r\) different from the given value is being used, M1 can still be allowed providing \(|r| < 1\).