C1 June 2016 Q9

EdexcelOld spec11 marksSequences & Series

9. On John’s 10th birthday he received the first of an annual birthday gift of money from his uncle. This first gift was £60 and on each subsequent birthday the gift was £15 more than the year before. The amounts of these gifts form an arithmetic sequence.

(a) Show that, immediately after his 12th birthday, the total of these gifts was £225 (1)

(b) Find the amount that John received from his uncle as a birthday gift on his 18th birthday. (2)

(c) Find the total of these birthday gifts that John had received from his uncle up to and including his 21st birthday. (3)

When John had received \(n\) of these birthday gifts, the total money that he had received from these gifts was £3375

(d) Show that \(n^2 + 7n = 25\times 18\) (3)

(e) Find the value of \(n\), when he had received £3375 in total, and so determine John’s age at this time. (2)

Mark scheme (a)

John; arithmetic series, \(a = 60,\ d = 15\).

Scheme	Marks
\(60 + 75 + 90 = 225\)* or \(S_3 = \dfrac{3}{2}\left(120 + (3 - 1)(15)\right) = 225\)*	B1 *
	(1)

Notes

B1 *: Finds and adds the first 3 terms or uses sum of 3 terms of an AP and obtains the printed answer, with no errors.

Beware:
The 12^th term of the sequence is 225 also so look out for \(60 + (12 - 1)\times 15 = 225\). This is B0.

Mark scheme (b)

Scheme	Marks
\(t_9 = 60 + (n - 1)15 = (\text{£})180\)	M1 A1
	(2)

Notes

M1: Uses \(60 + (n - 1)15\) with \(n = 8\) or 9

A1: (£)180

Listing:
M1: Uses \(a = 60\) and \(d = 15\) to select the 8^th or 9^th term (allow arithmetic slips)
A1: (£)180
(Special case (£)165 only scores M1A0)

Mark scheme (c)

Scheme	Marks
\(S_n = \dfrac{n}{2}\left(120 + (n - 1)(15)\right)\) or \(S_n = \dfrac{n}{2}\left(60 + 60 + (n - 1)(15)\right)\)	M1
\(S_n = \dfrac{12}{2}\left(120 + (12 - 1)(15)\right)\)	A1
\(= (\text{£})1710\)	A1
	(3)

Notes

M1: Uses correct formula for sum of \(n\) terms with \(a = 60\) and \(d = 15\) (must be a correct formula but ignore the value they use for \(n\) or could be in terms of \(n\))

A1: Correct numerical expression

A1: cao

Listing:
M1: Uses \(a = 60\) and \(d = 15\) and finds the sum of at least 12 terms (allow arithmetic slips)
A2: (£)1710

Mark scheme (d)

Scheme	Marks
\(3375 = \dfrac{n}{2}\left(120 + (n - 1)(15)\right)\)	M1
\(6750 = 15n\left(8 + (n - 1)\right) \Rightarrow 15n^2 + 105n = 6750\)	A1
\(n^2 + 7n = 25\times 18\)*	A1*
	(3)

Notes

M1: Uses correct formula for sum of \(n\) terms with \(a = 60\), \(d = 15\) and puts = 3375

A1: Correct three term quadratic. E.g. \(6750 = 105n + 15n^2,\ \ 3375 = \dfrac{15}{2}n^2 + \dfrac{105}{2}n\)
This may be implied by equations such as \(6750 = 15n(n + 7)\) or \(3375 = \dfrac{15}{2}\left(n^2 + 7n\right)\)

A1*: Achieves the printed answer with no errors but must see the 450 or 450 in factorised form or e.g. 6750, 3375 in factorised form i.e. an intermediate step.

Mark scheme (e)

Scheme	Marks
\(n = 18 \Rightarrow\) Aged 27	M1 A1
	(2)
	(11 marks)

Notes

M1: Attempts to solve the given quadratic or states \(n = 18\)

A1: Age = 27 or just 27

Age = 27 only scores both marks (i.e. \(n = 18\) need not be seen)

Note that (e) is not hence so allow valid attempts to solve the given equation for M1

\(n\)	1	2	3	4	5	6	7	8	9
\(u_n\)	60	75	90	105	120	135	150	165	180
\(S_n\)	60	135	225	330	450	585	735	900	1080
Age	10	11	12	13	14	15	16	17	18

\(n\)	10	11	12	13	14	15	16	17	18
\(u_n\)	195	210	225	240	255	270	285	300	315
\(S_n\)	1275	1485	1710	1950	2205	2475	2760	3060	3375
Age	19	20	21	22	23	24	25	26	27