C2 June 2016 Q1

Notes

(a) Way 2

(a) Way 3

Allow invisible brackets around fractions throughout all parts of this question.

M1 There are three possible methods as described above.

A1 Note that this is a ”show that” question with a printed answer.
In Way 1 this mark usually requires \(a = p/q\) where \(p\) and \(q\) may be unsimplified brackets from the formula (or could be 11200/175 for example) as an intermediate step before the conclusion \(a = 64\). Exceptions include \(a = 175/4 * 256/175\) i.e. multiplication by reciprocal rather than division or \(175 = 175a/64\) followed by the obvious \(a = 64\) These also get A1
In “reverse” methods such as Way 3 we need a conclusion “so \(a = 64\)” or some implication that their argument is reversible. Also a conclusion can be implied from a preamble, eg: “If I assume \(a = 64\) then find \(S\)= 175 as given this implies \(a = 64\) as required”
This is a show that question and there should be no loss of accuracy.
In all the methods if decimals are used there should not be rounding.
If 0.68359375 appears this is correct. If it is rounded it would not give the exact answer.
\(64(1 - 0.31640625)\) or 43.75 are each correct – if they are rounded then treat this as incorrect
e.g. Way 3: “43.75/0.25 = 175 so \(a\) = 64 is A1” but “43/0.25 = 175 so \(a\) = 64 is A0” and “44/0.25 = 175 so a = 64 is A0”
Yet another variant on Way 3: take a=64 then find the next 3 terms as 48, 36, 27 then add 64+48+36+27 to get 175. Again need conclusion that \(a = 64\) or some implication that their argument is reversible. Otherwise M1 A0

Notes

M1 \(S_\infty = \dfrac{64}{1 - \frac{3}{4}}\) or \(\dfrac{(\text{their } a \text{ found in part } (a))}{1 - \frac{3}{4}}\)

A1 256 cao

Notes

M1: Writes down either “64”\(\left(\dfrac{3}{4}\right)^8\) or awrt 6.4 or “64”\(\left(\dfrac{3}{4}\right)^9\) or awrt 4.8, using \(a = 64\) or their \(a\)

dM1: A correct expression for the difference (i.e. \(\pm(T_9 - T_{10})\)) using \(a = 64\) or their \(a\).

NB Using Sum of 10 terms minus Sum of 9 terms is NOT a misread Scores M0M0A0

M1 Can be implied. Writes down either \(64\left(\dfrac{3}{4}\right)^8\) or \(64\left(\dfrac{3}{4}\right)^9\), using \(a = 64\) (or their \(a\) found in part (a)).

Note Ignore candidate’s labelling of terms.

Note \(64\left(\dfrac{3}{4}\right)^8 = 6.407226563\ldots\) and \(64\left(\dfrac{3}{4}\right)^9 = 4.805419922\ldots\)

dM1 This is dependent on previous M mark and can be implied. Either \(64\left(\dfrac{3}{4}\right)^8 - 64\left(\dfrac{3}{4}\right)^9\) or \(64\left(\dfrac{3}{4}\right)^9 - 64\left(\dfrac{3}{4}\right)^8\) or awrt 6.4 – awrt 4.8 , using \(a = 64\) (or their \(a\) from part (a))

Note 1^st M1 and 2^nd M1 can be implied by the value of their difference \(=\) "their \(a\) found in part (a)" \(\times \dfrac{3^8}{4^9} \approx \dfrac{\text{"their } a \text{ found in part (a)"}}{40}\)

Note Either \(64\left(\dfrac{3}{4}\right)^9 - 64\left(\dfrac{3}{4}\right)^{10}\) or \(64\left(\dfrac{3}{4}\right)^{10} - 64\left(\dfrac{3}{4}\right)^9\) is 1^st M1, 2^nd M0.

A1 1.602 or \(-1.602\) cao (This answer with no working is M1M1A1) But 1.6 with no working is M0M0A0

Note \(\left\{D = \dfrac{1}{4}T_9 \Rightarrow\right\}\ D = \dfrac{1}{4}(64)\left(\dfrac{3}{4}\right)^8\) is 1^st M1, 2^nd M1

Special case Obtains awrt 6.4, then obtains awrt 4.8 but rounds to 6 – 5 when subtracting – award M1M1A0