C2 June 2017 Q9

EdexcelOld spec12 marksSequences & Series

9. The first three terms of a geometric sequence are\[7k - 5, \ 5k - 7, \ 2k + 10\]where \(k\) is a constant.

(a) Show that \(11k^2 - 130k + 99 = 0\) (4)

Given that \(k\) is not an integer,

(b) show that \(k = \dfrac{9}{11}\) (2)

For this value of \(k\),

(c)

(i) evaluate the fourth term of the sequence, giving your answer as an exact fraction,

(ii) evaluate the sum of the first ten terms of the sequence.

(6)

Mark scheme (a)

Scheme	Marks
\(a = 7k - 5,\ ar = 5k - 7\) and \(ar^2 = 2k + 10\)	B1
(So \(r =\) ) \(\dfrac{5k - 7}{7k - 5} = \dfrac{2k + 10}{5k - 7}\) or \((7k - 5)(2k + 10) = (5k - 7)^2\) or equivalent	M1
See \((5k - 7)^2 = 25k^2 - 70k + 49\)	M1
\(14k^2 + 60k - 50 = 25k^2 - 70k + 49 \to 11k^2 - 130k + 99 = 0\) *	A1cso *
	(4)

Notes

Mark parts (a) and (b) together

B1: Correct statement (needs all three terms)– this may be omitted and implied by correct statement in \(k\) only, as candidates may use geometric mean, or may use ratio of terms being equal and give a correct line 2 without line 1. (This would earn the B1M1 immediately)

M1: Valid Attempt to eliminate \(a\) and \(r\) and to obtain equation in \(k\) only

M1: Correct expansion of \((5k - 7)^2 = 25k^2 - 70k + 49\) - may have four terms \((5k - 7)^2 = 25k^2 - 35k - 35k + 49\)

A1cso: No incorrect work seen. The printed answer is obtained including “=0”.

Mark scheme (b)

Scheme	Marks
\((k - 11)(11k - 9)\) so \(k =\)	M1
\(k = 9/11\) only* (after rejecting 11)	A1*
N.B. Special case \(k = 9/11\) can be verified in (b) (1 mark only) \(11 \times \left(\dfrac{9}{11}\right)^2 - 130 \times \left(\dfrac{9}{11}\right) + 99 = \dfrac{81}{11} - \dfrac{1170}{11} + \dfrac{1089}{11} = 0\) M1A0
	(2)

Notes

M1: Attempt to solve quadratic by usual methods (factorisation, completion of square or formula – see notes at start of mark scheme) or see 9/11 substituted and given as “=0” for M1A0

A1*: 9/11 only and 11 should be seen and rejected. Accept 9/11 underlined or \(k = 9/11\) written on following line.
Alternatively \((k - 11)\) may be seen in the factorisation and a statement ‘\(k\) not integer’ given with \(k = 9/11\) stated.

Mark scheme (c)

Scheme	Marks
\(a = \dfrac{8}{11}\)	B1
\(\dfrac{5 \times \frac{9}{11} - 7}{7 \times \frac{9}{11} - 5}\) or \(\dfrac{2 \times \frac{9}{11} + 10}{5 \times \frac{9}{11} - 7}\) so \(r = -4\)	B1
(i) Fourth term \(= ar^3 = -\dfrac{512}{11}\)	M1A1
(ii) \(S_{10} = \dfrac{a(1 - r^{10})}{(1 - r)} = \dfrac{\frac{8}{11}\left(1 - (-4)^{10}\right)}{(1 - (-4))} = -152520\)	M1A1
	(6)
	[12]

Notes

Mark parts (i) and (ii) together

B1: \(a = \dfrac{8}{11}\) or any equivalent (If not stated explicitly or used in formula may be implied by correct answer to (ii))

B1: Substitutes \(k = 9/11\) completely and obtain \(r = -4\) (If not stated explicitly, may be implied by correct answer to (i) or (ii))

(i) M1: Use of correct formula with \(n = 4\) \(a\) and/or \(r\) may still be in terms of \(k\) or uses \((2k + 10) \times r\). May assume \(r = k\).
A1: Correct exact answer

(ii) M1: Use of correct formula with \(n = 10\) \(a\) and/or \(r\) may still be in terms of \(k\) May assume \(r = k\) A1 : \(-152520\) cao

NB Correct formula with negative sign in numerator followed by the incorrect \((8/11)(1 + 4^{10})/(1 - (-4))\) usually found equal to 152520.2909 with no negative sign can be allowed M1A0 but if the incorrect numerical expression appears on its own with no formula then M0A0

Listing terms can get: B1 (first term) B1 M1A1 (implied by correct 4^th term) M1A1 (implied by \(-152520\))