C1 June 2012 Q6

EdexcelOld spec10 marksSequences & Series

6. A boy saves some money over a period of 60 weeks. He saves 10p in week 1, 15p in week 2, 20p in week 3 and so on until week 60. His weekly savings form an arithmetic sequence.

(a) Find how much he saves in week 15 (2)

(b) Calculate the total amount he saves over the 60 week period. (3)

The boy’s sister also saves some money each week over a period of \(m\) weeks. She saves 10p in week 1, 20p in week 2, 30p in week 3 and so on so that her weekly savings form an arithmetic sequence. She saves a total of £63 in the \(m\) weeks.

(d) Hence write down the value of \(m\). (1)

Mark scheme (a)

Scheme	Marks
Boy’s Sequence: 10, 15, 20, 25, ...
\(\{a = 10,\ d = 5 \Rightarrow T_{15} =\}\ a + 14d = 10 + 14(5);\ = 80\) or \(0.1 + 14(0.05);\ =\) £0.80	M1; A1
	(2)

Notes

M1: for using the formula \(a + 14d\) with either \(a\) or \(d\) correct.

A1: for 80 or 80p or £0.80 or £0.80p and apply ISW. Otherwise, £80 or 0.80 or 0.80p would be A0.

Award M0 if candidate applies \(a + 59d\).
Listing the first 15 terms and highlighting that the 15^th term is 80 or listing 15 terms with the final 15^th term aligned with 80 will then be awarded all two marks of M1A1.
Writing down 80 with no working is M1A1.

Mark scheme (b)

Scheme	Marks
\(\{S_{60} =\}\ \dfrac{60}{2}\left[2(10) + 59(5)\right]\)	M1 A1
\(= 30(315) = 9450\) or £94.50	A1
	(3)

Notes

M1: for use of correct \(\dfrac{60}{2}\left[2(10) + 59(5)\right]\) or \(\dfrac{15}{2}\left(2(10) + 14(5)\right)\) with \(a = 10,\ d = 5\) and \(n = 60\) or \(a = 10,\ d = 5\) and \(n = 15\).
If a candidate uses \(\dfrac{n}{2}(a + l)\) with \(n = 60\) or 15, there must be a full method of finding or stating \(l\) as either \(a + 59d\ (= 305)\) or \(a + 14d\ (= 80)\), respectively.

1^st A1: for a correct expression for \(S_{60}\). ie. \(\dfrac{60}{2}\left[2(10) + 59(5)\right]\) or \(\dfrac{60}{2}\left[2(0.1) + 59(0.05)\right]\) or \(\dfrac{60}{2}\left[10 + 305\right]\) or \(\dfrac{60}{2}\left[0.10 + 3.05\right]\). This mark can be implied by later working.

2^nd A1: for 9450 or 9450p or £94.50 and apply ISW. Otherwise, £9450 or 94.50 without £ sign is A0.

Note: the bracketing error of \(\dfrac{60}{2}2(10) + 59(5)\) is A0 unless recovered from later working.
Adding together the first 60 terms to obtain 9450 will then be awarded all three marks of M1A1A1.

Mark scheme (c)

Scheme	Marks
Boy’s Sister’s Sequence: 10, 20, 30, 40, ...
\(\{a = 10,\ d = 10 \Rightarrow S_m =\}\ \dfrac{m}{2}\left(2(10) + (m - 1)(10)\right)\) \(\left(\text{or } \dfrac{m}{2}\times 10(m + 1) \text{ or } 5m(m + 1)\right)\)	M1 A1
63 or \(6300 = \dfrac{m}{2}\left(2(10) + (m - 1)(10)\right)\)	dM1
\(6300 = \dfrac{m}{2}(10)(m + 1)\) or \(12600 = 10m(m + 1)\) \(1260 = m(m + 1)\) \(35\times 36 = m(m + 1) \quad (*)\)	A1 cso
	(4)

Notes

1^st M1: for correct use of \(S_m\) formula with one of \(a\) or \(d\) correct.

1^st A1: for a correct expression for \(S_m\). Eg: \(\dfrac{m}{2}\left(2(10) + (m - 1)(10)\right)\) or \(\dfrac{m}{2}\times 10(m + 1)\) or \(5m(m + 1)\)

2^nd M1: for forming a suitable equation using 63 or 6300 and their \(S_m\). Dependent on 1^st M1.

2^nd A1cso: for reaching the printed result with no incorrect working seen.
Long multiplication is not necessary for the final accuracy mark.

Note: \(\dfrac{m}{2}\left(2(10) + (m - 1)(10)\right) = 630\) and not either 6300 or 63 is dM0.Beware: Some candidates will try and fudge the result given on the question paper.Notes for awarding 2^nd A1
Going from \(m(m + 1) = 1260\) straight to \(m(m + 1) = 35\times 36\) is 2^nd A1.
Going from \(m(m + 1) =\) some factor decomposition of 6300 straight to \(m(m + 1) = 35\times 36\) is 2^nd A1.
Going from \(10m(m + 1) = 12600\) straight to \(m(m + 1) = 35\times 36\) is 2^nd A0.
Going from \(m(m + 1) = \dfrac{6300}{5}\) straight to \(m(m + 1) = 35\times 36\) is 2^nd A0.

Alternative: working in an different letter, say n or p.

M1A1: for \(\dfrac{n}{2}\left(2(10) + (n - 1)(10)\right)\) (although mixing letters eg. \(\dfrac{n}{2}\left(2(10) + (m - 1)(10)\right)\) is M0A0).

dM1: for 63 or \(6300 = \dfrac{n}{2}\left(2(10) + (n - 1)(10)\right)\)

Leading to \(6300 = \dfrac{n}{2}(10)(n + 1) \Rightarrow 1260 = n(n + 1) \Rightarrow 35\times 36 = n(n + 1)\)
The candidate then needs to write either \(35\times 36 = m(m + 1)\) or \(m \equiv n\) or \(m = n\) to gain the final A1.

Scheme	Marks
\(\{m =\}\ 35\)	B1
	(1)
	(10 marks)