C1 June 2005 Q9

EdexcelOld spec13 marksSequences & Series

9. An arithmetic series has first term \(a\) and common difference \(d\).

(a) Prove that the sum of the first \(n\) terms of the series is\[\tfrac{1}{2}n[2a + (n - 1)d].\] (4)

Sean repays a loan over a period of \(n\) months. His monthly repayments form an arithmetic sequence.

He repays £149 in the first month, £147 in the second month, £145 in the third month, and so on. He makes his final repayment in the \(n\)th month, where \(n > 21\).

(b) Find the amount Sean repays in the 21st month. (2)

Over the \(n\) months, he repays a total of £5000.

(d) Solve the equation in part (c). (3)

(e) State, with a reason, which of the solutions to the equation in part (c) is not a sensible solution to the repayment problem. (1)

Mark scheme (a)

Scheme	Marks
\((S =)\,a + (a + d) + \ldots \quad \ldots + [a + (n - 1)d]\)	B1
\((S =)\,[a + (n - 1)d] + \ldots \quad \ldots + a\)	M1
\(\left.\begin{aligned} 2S &= [2a + (n - 1)d] + \ldots \quad \ldots + [2a + (n - 1)d] \\ 2S &= n[2a + (n - 1)d] \end{aligned}\right\}\) either	dM1
\(S = \dfrac{n}{2}[2a + (n - 1)d]\)	A1 c.s.o
	(4)

Notes

(a) B1 requires at least 3 terms, must include first and last terms, an adjacent term dots and + signs.
1^st M1 for reversing series. Must be arithmetic with \(a\), \(d\) (or \(a\), \(l\)) and \(n\).
2^nd dM1 for adding, must have \(2S\) and be a genuine attempt. Either line is sufficient. Dependent on 1^st M1
(NB Allow first 3 marks for use of \(l\) for last term but as given for final mark)

Scheme	Marks
\((a = 149,\ d = -2)\) \(u_{21} = 149 + 20(-2) = \)£109	M1 A1
	(2)

Mark scheme (c)

Scheme	Marks
\(S_n = \dfrac{n}{2}[2\times 149 + (n - 1)(-2)]\) \(\left(= n(150 - n)\right)\)	M1 A1
\(S_n = 5000 \Rightarrow n^2 - 150n + 5000 = 0\) (*)	A1 c.s.o
	(3)

Notes

(c) M1 for using their \(a, d\) in \(S_n\) A1 any correct expression
A1cso for putting \(S_n = 5000\) and simplifying to given expression. No wrong work

Mark scheme (d)

Scheme	Marks
\((n - 100)(n - 50) = 0\)	M1
\(n = 50\) or 100	A2/1/0
	(3)

Notes

(d) M1 Attempt to solve leading to \(n = \ldots\)
A2/1/0 Give A1A0 for 1 correct value and A1A1 for both correct

Mark scheme (e)

Scheme	Marks
\(u_{100} < 0 \qquad \therefore n = 100\) not sensible	B1 f.t.
	(1)
	(13 marks)

Notes

(e) B1 f.t. Must mention 100 and state \(u_{100} < 0\) (or loan paid or equivalent)
If giving f.t. then must have \(n \geqslant 76\).