C1 January 2006 Q7

EdexcelOld spec13 marksSequences & Series

7. On Alice’s 11th birthday she started to receive an annual allowance. The first annual allowance was £500 and on each following birthday the allowance was increased by £200.

(a) Show that, immediately after her 12th birthday, the total of the allowances that Alice had received was £1200. (1)

(b) Find the amount of Alice’s annual allowance on her 18th birthday. (2)

When the total of the allowances that Alice had received reached £32000 the allowance stopped.

(d) Find how old Alice was when she received her last allowance. (7)

Scheme	Marks
\(500 + (500 + 200) = 1200\) or \(S_2 = \dfrac{1}{2}2\{1000 + 200\} = 1200\) (*)	B1
	(1)

Mark scheme (b)

Scheme	Marks
Using \(a = 500,\ d = 200\) with \(n = 7, 8\) or 9 \(a + (n - 1)d\) or “listing”	M1
\(500 + (7\times 200) = (\)£\()1900\)	A1
	(2)

Notes

(b) Correct answer with no working: Allow both marks.

Mark scheme (c)

Scheme	Marks
Using \(\dfrac{1}{2}n\{2a + (n - 1)d\}\) or \(\dfrac{1}{2}n\{a + l\}\), or listing and “summing” terms	M1
\(S_8 = \dfrac{1}{2}8\{2\times 500 + 7\times 200\}\) or \(S_8 = \dfrac{1}{2}8\{500 + 1900\}\), or all terms in list correct	A1
\(= (\)£\()\ 9600\)	A1
	(3)

Notes

(c) Some working must be seen to score marks:
Minimum working: \(500 + 700 + 900 + \ldots (+\,1900) = \ldots\) scores M1 (A1).

Mark scheme (d)

Scheme	Marks
\(\dfrac{1}{2}n\{2\times 500 + (n - 1)\times 200\} = 32000\) M1: General \(S_n\), equated to 32000	M1 A1
\(n^2 + 4n - 320 = 0\) (or equiv.) M1: Simplify to 3 term quadratic	M1 A1
\((n + 20)(n - 16) = 0 \quad n = \ldots\) M1: Attempt to solve 3 t.q.	M1
\(n = 16\), Age is 26	A1cso,A1cso
	(7)
	(13 marks)

Notes

(d) Allow \(\geqslant\) or \(>\) throughout, apart from “Age 26”.

A common misread here is 3200. This gives \(n = 4\) and age 14, and can score M1 A0 M1 A0 M1 A1 A1 with the usual misread rule.

Alternative: (Listing sums)

(500, 1200, 2100, 3200, 4500, 6000, 7700, 9600,) 11700, 14000, 16500, 19200, 22100, 25200, 28500, 32000.

Scheme	Marks
List at least up to 32000	M3
All values correct	A2
\(n = 16\) (perhaps implied by age)	A1cso
Age 26	A1cso

If there is a mistake in the list, e.g. 16^th sum = 32100, possible marks are: M3 A0 A0 A0

Alternative: (Trial and improvement)

Scheme	Marks
Use of \(S_n\) formula with \(n = 16\) (and perhaps other values)	M3
Accurately achieving 32000 for \(n = 16\)	A3
Age 26	A1