C2 June 2013 (R) Q5

EdexcelOld spec11 marksSequences & Series

5. The first three terms of a geometric series are \(4p\), \((3p + 15)\) and \((5p + 20)\) respectively, where \(p\) is a positive constant.

(a) Show that \(11p^2 - 10p - 225 = 0\) (4)

(b) Hence show that \(p = 5\) (2)

(d) Find the sum of the first ten terms of the series, giving your answer to the nearest integer. (3)

Mark scheme (a)

Scheme	Marks
\(a = 4p,\ ar = (3p + 15)\) and \(ar^2 = 5p + 20\)	B1
(So \(r =\)) \(\dfrac{5p + 20}{3p + 15} = \dfrac{3p + 15}{4p}\) or \(4p(5p + 20) = (3p + 15)^2\) or equivalent	M1
See \((3p + 15)^2 = 9p^2 + 90p + 225\)	M1
\(20p^2 + 80p = 9p^2 + 90p + 225 \to 11p^2 - 10p - 225 = 0\ \ *\)	A1 *
	(4)

Notes

B1: Correct statement (needs all three terms)– this may be omitted and implied by correct statement in \(p\) 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 \(p\) only

M1: Correct expansion of \((3p + 15)^2 = 9p^2 + 90p + 225\)

A1cso: No incorrect work seen. The printed answer is obtained.

NB Those who show \(p = 5\) in part (a) obtain no credit for this

Mark scheme (b)

Scheme	Marks
\((p - 5)(11p + 45)\) so \(p =\)	M1
\(p = 5\) only ( after rejecting \(-45/11\) )	A1
N.B. Special case \(p = 5\) can be verified in (b) (1 mark only) \(11 \times 5^2 - 10 \times 5 - 225 = 275 - 50 - 225 = 0\) M1A0
	(2)

Notes

M1: Attempt to solve quadratic by usual methods (factorisation, completion of square or formula) Must appear in part (b) – not part (a)

A1: 5 only and -45/11 should be seen and rejected or \((11p + 45)\) seen and statement \(p > 0\)

Mark scheme (c)

Scheme	Marks
\(\dfrac{3 \times 5 + 15}{4 \times 5}\) or \(\dfrac{5 \times 5 + 20}{3 \times 5 + 15}\)	M1
\(r = \dfrac{3}{2}\)	A1
	(2)

Notes

M1: Substitutes \(p = 5\) completely and attempt ratio (correct way up)

A1: 1.5 or any equivalent

Mark scheme (d)

Scheme	Marks
\(S_{10} = \dfrac{20\left(1 - \left(\text{"}\frac{3}{2}\text{"}\right)^{10}\right)}{\left(1 - \text{"}\frac{3}{2}\text{"}\right)}\)	M1A1ft
\((= 2266.601568\ldots) = 2267\)	A1
	(3)
	Total 11

Notes

M1: Use of correct formula with \(n = 10\) \(a\) and/or \(r\) may still be in terms of \(p\)

A1ft: Correct expression ft on their \(r\) only – must have \(a = 20\) and power = 10 here

A1 2267 (accept awrt 2267)