C2 January 2011 Q3

Notes

B1: for both \(ar = 750\) and \(ar^4 = -6\) (may be implied from later working in either (a) or (b)).

M1: for eliminating a by either dividing \(ar^4 = -6\) by \(ar = 750\) or dividing \(ar = 750\) by \(ar^4 = -6\), to achieve an equation in \(r^3\) or \(\dfrac{1}{r^3}\). Note that \(r^4 - r = -\dfrac{6}{750}\) is M0.

Note also that any of \(r^3 = \dfrac{-6}{750}\) or \(r^3 = \dfrac{750}{-6}\ \{= -125\}\) or \(\dfrac{1}{r^3} = \dfrac{-6}{750}\) or \(\dfrac{1}{r^3} = \dfrac{750}{-6}\ \{= -125\}\) are fine for the award of M1.

SC: \(ar^\alpha = 750\) and \(ar^\beta = -6\) leading to \(r^\delta = \dfrac{-6}{750}\) or \(r^\delta = \dfrac{750}{-6}\ \{= -125\}\) or \(\dfrac{1}{r^\delta} = \dfrac{-6}{750}\) or \(\dfrac{1}{r^\delta} = \dfrac{750}{-6}\ \{= -125\}\) where \(\delta = \beta - \alpha\) and \(\delta \geqslant 2\) are fine for the award of M1.

SC: \(ar^2 = 750\) and \(ar^5 = -6\) leading to \(r = -\tfrac{1}{5}\) scores B0M1A1.

Notes

M1 for inserting their \(r\) into either of their original correct equations of either \(ar = 750\) or \(\{a =\}\ \dfrac{750}{r}\) or \(ar^4 = -6\) or \(\{a =\}\ \dfrac{-6}{r^4}\) – in both a and r. No slips allowed here for M1.

A1 for either \(a = -3750\) or \(a\) equal to the correct follow through result expressed either as an exact integer, or a fraction in the form \(\dfrac{c}{d}\) where both \(c\) and \(d\) are integers, or correct to awrt 1 dp.

Notes

M1 for applying \(\dfrac{a}{1 - r}\) correctly (only a slip in substituting \(r\) is allowed) using both their \(a\) and their \(|r| \lt 1\). Eg. \(\dfrac{-3750}{1 - -0.2}\).    A1 for \(-3125\)

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