C1 June 2015 Q9

EdexcelOld spec7 marksSequences & Series

9. Jess started work 20 years ago. In year 1 her annual salary was £17000. Her annual salary increased by £1500 each year, so that her annual salary in year 2 was £18500, in year 3 it was £20000 and so on, forming an arithmetic sequence. This continued until she reached her maximum annual salary of £32000 in year \(k\). Her annual salary then remained at £32000.

(a) Find the value of the constant \(k\). (2)

(b) Calculate the total amount that Jess has earned in the 20 years. (5)

Mark scheme (a)

Scheme	Marks
\(32000 = 17000 + (k - 1)\times 1500 \Rightarrow k = \ldots\)	M1
\((k =)\ 11\)	A1
	(2)

Notes

M1: Use of 32000 with a correct formula in an attempt to find \(k\). A correct formula could be implied by a correct answer.

A1: Cso (Allow \(n = 11\))

Accept correct answer only.

32000 = 17000 + 1500\(k \Rightarrow k = 10\) is M0A0 (wrong formula)
\(\dfrac{32000 - 17000}{1500} = 10 \therefore k = 11\) is M1A1 (correct formula implied)

Listing: All terms must be listed up to 32000 and 11 correctly identified. A solution that scores 2 if fully correct and 0 otherwise.

Mark scheme (b)

Scheme	Marks
M1: \(S = \tfrac{k}{2}\left(2\times 17000 + (k - 1)\times 1500\right)\) or \(\tfrac{k}{2}(17000 + 32000)\) \(S = \tfrac{k-1}{2}\left(2\times 17000 + (k - 2)\times 1500\right)\) or \(\tfrac{k-1}{2}(17000 + 30500)\) A1: \(S = \tfrac{11}{2}(2\times 17000 + 10\times 1500)\) or \(\tfrac{11}{2}(17000 + 32000)\) \(S = \tfrac{10}{2}(2\times 17000 + 9\times 1500)\) or \(\tfrac{10}{2}(17000 + 30500)\) (= 269 500 or 237 500)	M1A1
\(32000\times\alpha\)	M1
288 000 + 269 500 = 557 500 or 320 000 + 237 500 = 557 500	ddM1A1
	(5)
	(7 marks)

Notes

M1: Use of correct sum formula with their integer \(n = k\) or \(k - 1\) from part (a) where \(3 < k < 20\) and \(a = 17000\) and \(d = 1500\). See below for special case for using \(n = 20\).

A1: Any correct un-simplified numerical expression with \(n = 11\) or \(n = 10\)

M1: \(32000\times\alpha\) where \(\alpha\) is an integer and \(3 < \alpha < 18\)

ddM1: Attempts to add their two values. It is dependent upon the two previous M’s being scored and must be the sum of 20 terms i.e. \(\alpha + k = 20\)

A1: 557 500

Special Case: If they just find S₂₀ (£625 000) in (b) score the first M1 otherwise apply the scheme.

Listing:

\(n\)	1	2	3	4	5	6	7	8	9	10
\(u_n\)	17000	18500	20000	21500	23000	24500	26000	27500	29000	30500
\(n\)	11	12	13	14	15	16	17	18	19	20
\(u_n\)	32000	32000	32000	32000	32000	32000	32000	32000	32000	32000

Look for a sum before awarding marks. Award the M’s as above then A2 for 557 500
If they sum the ‘parts’ separately then apply the scheme.