C2 June 2011 Q6

Notes

M1: for eliminating a by eg. \(192r = 144\) or by either dividing \(ar^2 = 144\) by \(ar = 192\) or dividing \(ar = 192\) by \(ar^2 = 144\), to achieve an equation in \(r\) or \(\dfrac{1}{r}\). Note that \(r^2 - r = \dfrac{144}{192}\) is M0.

Note also that any of \(r = \dfrac{144}{192}\) or \(r = \dfrac{192}{144}\left\{= \dfrac{4}{3}\right\}\) or \(\dfrac{1}{r} = \dfrac{192}{144}\) or \(\dfrac{1}{r} = \dfrac{144}{192}\) are fine for the award of M1. Note: A candidate just writing \(r = \dfrac{144}{192}\) with no reference to \(a\) can also get the method mark.

Note: \(ar^2 = 192\) and \(ar^3 = 144\) leading to \(r = \tfrac{3}{4}\) scores M1A1. This is because \(r\) is the ratio between any two consecutive terms. These candidates, however, will usually be penalised in part (b).

Notes

M1 for inserting their \(r\) into either of the correct equations of either \(ar = 192\) or \(\{a =\}\ \dfrac{192}{r}\) or \(ar^2 = 144\) or \(\{a =\}\ \dfrac{144}{r^2}\). No slips allowed here for M1.

M1: can also be awarded for writing down \(144 = a\left(\dfrac{192}{a}\right)^2\)

A1 for \(a = 256\) only. Note 256 from any working scores M1A1.

Note: Some candidates incorrectly confuse notation to give \(r = \tfrac{4}{3}\) or 1.33 in part (a) (getting M1A0). In part (b), they recover to write \(a = 192 \times \tfrac{4}{3}\) for M1 and then 256 for A1.

Notes

M1: for applying \(\dfrac{a}{1 - r}\) correctly (no slips allowed!) using both their \(a\) and their \(r\), where \(|r| \lt 1\).

A1: for 1024, cao.

In parts (a) or (b) or (c), the correct answer with no working scores full marks.

Notes

1^st M1: For applying \(\mathrm{S}_n\) with their \(a\) and either “the letter \(r\)” or their \(r\) and “uses” 1000.

2^nd M1: For isolating \(+(r)^n\) and not \((ar)^n\), (eg. \((192)^n\)) as the subject of an equation or inequality. \(+(r)^n\) must be derived from the \(\mathrm{S}_n\) formula.

3^rd M1: For applying the power law to \(\lambda^k = \mu\) to give \(k\log\lambda = \log\mu\) oe. where \(\lambda, \mu \gt 0\).
or 3^rd M1: For solving \(\lambda^k = \mu\) to give \(k = \log_\lambda \mu\), where \(\lambda, \mu \gt 0\).

A1: cso If a candidate uses inequalities, a fully correct method with inequalities is required here. So, an incorrect inequality statement at any stage in a candidate’s working for this part loses this mark.
Note: Some candidates do not realise that the direction of the inequality is reversed in the final line of their solution.

Or A1: cso Note a candidate can achieve full marks here if they do not use inequalities. So, if a candidate uses equations rather than inequalities in their working then they need to state in the final line of their working that \(n = 13.04\) (truncated) or \(n = \text{awrt } 13.05 \Rightarrow n = 14\) for A1.

\(n = 14\) from no working gets SC: M0M0M1A1.

A method of \(\mathrm{T}_n \gt 1000 \Rightarrow 256(0.75)^{n - 1} \gt 1000\) can score M0M0M1A0 for a correct application of the power law of logarithms.

Trial & Improvement Method

For \(a = 256\) and \(r = 0.75\), apply the following scheme:

Note: A similar scheme would apply for T&I for candidates using their \(a\) and their \(r\). So,...
1^st M1: For attempting to find one of the correct \(\mathrm{S}_n\)’s either side (but next to) 1000.
2^nd M1: For one of these \(\mathrm{S}_n\)’s correct for their \(a\) and their \(r\). (You may need to get your calculators out!)
3^rd M1: For attempting to find both of the correct \(\mathrm{S}_n\)’s either side (but next to) 1000.
A1: Cannot be gained for wrong \(a\) and/or \(r\).

Trial & Improvement Cumulative Approach:
A similar scheme to T&I will be applied here:
1^st M1: For getting as far as the cumulative sum of 13 terms. 2^nd M1: (1) \(\mathrm{S}_{13} = \text{awrt } 999.7\) or truncated 999. 3^rd M1: For getting as far as the cumulative sum to 14 terms. Also at this stage \(\mathrm{S}_{13} \lt 1000\) and \(\mathrm{S}_{14} \gt 1000\). A1: BOTH (1) \(\mathrm{S}_{13} = \text{awrt } 999.7\) or truncated 999 AND (2) \(\mathrm{S}_{14} = \text{awrt } 1005.8\) or truncated 1005 AND \(n = 14\).

Trial & Improvement Method: for \((0.75)^n \lt \tfrac{6}{256} = 0.0234375\)
3^rd M1: For evidence of examining both \(n = 13\) and \(n = 14\).
Eg: \((0.75)^{13}\ \{= 0.023757\ldots\}\) and \((0.75)^{14}\ \{= 0.0178179\ldots\}\)
A1: \(n = 14\)

Any misreads, \(\mathrm{S}_n \gt 10000\) etc, please escalate up to your Team Leader.