C2 June 2014 (R) Q2

EdexcelOld spec8 marksSequences & Series

2. A geometric series has first term \(a\), where \(a \neq 0\), and common ratio \(r\).
The sum to infinity of this series is 6 times the first term of the series.

(a) Show that \(r = \dfrac{5}{6}\) (2)

Given that the fourth term of this series is 62.5

(b) find the value of \(a\), (2)

(c) find the difference between the sum to infinity and the sum of the first 30 terms, giving your answer to 3 significant figures. (4)

Mark scheme (a)

Scheme	Marks
\(\mathrm{S}_\infty = 6a\)
\(\dfrac{a}{1 - r} = 6a\)	M1
\(\{\Rightarrow 1 = 6(1 - r) \Rightarrow\}\ r = \dfrac{5}{6}\) *	A1*
Allow verification e.g. \(\dfrac{a}{1 - r} = 6a \Rightarrow \dfrac{a}{1 - \frac{5}{6}} = 6a \Rightarrow \dfrac{a}{\frac{1}{6}} = 6a \Rightarrow 6a = 6a\)
	(2)

Notes

M1: Either \(\dfrac{a}{1 - r} = 6a\) or \(\dfrac{6a}{1 - r} = a\) or \(\dfrac{6}{1 - r} = 1\)

A1*: cso

Mark scheme (b)

Scheme	Marks
\(\left\{\mathrm{T}_4 = ar^3 = 62.5 \Rightarrow\right\}\ a\left(\dfrac{5}{6}\right)^3 = 62.5\)	M1
\(\Rightarrow a = 108\)	A1
	(2)

Notes

M1: \(a\left(\dfrac{5}{6}\right)^3 = 62.5\) (Correct statement using the 4^th term. Do not accept \(a\left(\dfrac{5}{6}\right)^4 = 62.5\))

A1: 108

Mark scheme (c)

Scheme	Marks
\(\mathrm{S}_\infty = 6(\text{their } a)\) or \(\dfrac{\text{their } a}{1 - \frac{5}{6}}\ \{= 648\}\)	M1
\(\{\mathrm{S}_{30} =\}\ \dfrac{108\left(1 - \left(\frac{5}{6}\right)^{30}\right)}{1 - \frac{5}{6}}\ \{= 645.2701573\ldots\}\)	M1 A1ft
\(\{\mathrm{S}_\infty - \mathrm{S}_{30}\} = 2.72984\ldots\) awrt 2.73	A1
	(4)
	Total 8

Notes

M1: Correct method to find \(\mathrm{S}_\infty\)

M1: \(\mathrm{S}_{30} = \dfrac{(\text{their } a)\left(1 - \left(\frac{5}{6}\right)^{30}\right)}{1 - \left(\frac{5}{6}\right)}\) (Condone invisible brackets around 5/6)

A1ft: Correct follow through expression (follow through their \(a\)). Do not condone invisible brackets around 5/6 unless their evaluation or final answer implies they were intended.

Alternative (c):

Difference \(= \dfrac{ar^{30}}{1 - r} = \dfrac{108\left(\frac{5}{6}\right)^{30}}{1 - \frac{5}{6}} = 2.72984\ldots\)
M1M1: For an attempt to apply \(\dfrac{ar^{30}}{1 - r}\).
A1ft: \(\dfrac{(\textit{their } a) \times r^{30}}{1 - r}\) with their ft \(a\).
A1: awrt 2.73