C2 June 2010 Q9

EdexcelOld spec10 marksSequences & Series

9. The adult population of a town is 25 000 at the end of Year 1.

A model predicts that the adult population of the town will increase by 3% each year, forming a geometric sequence.

(a) Show that the predicted adult population at the end of Year 2 is 25 750. (1)

(b) Write down the common ratio of the geometric sequence. (1)

The model predicts that Year \(N\) will be the first year in which the adult population of the town exceeds 40 000.

(d) Find the value of \(N\). (2)

At the end of each year, each member of the adult population of the town will give £1 to a charity fund.

Assuming the population model,

(e) find the total amount that will be given to the charity fund for the 10 years from the end of Year 1 to the end of Year 10, giving your answer to the nearest £1000. (3)

Scheme	Marks
(a) 25 000 × 1.03 = 25750 \(\left\{25000 + 750 = 25750,\ \text{or}\ \ 25000\dfrac{(1 - 0.03^2)}{1 - 0.03} = 25750\right\}\) (*)	B1
	(1)

Scheme	Marks
(b) \(r = 1.03\) Allow \(\dfrac{103}{100}\) or \(1\dfrac{3}{100}\) but no other alternatives	B1
	(1)

Mark scheme (c)

Scheme	Marks
(c) \(25000r^{N-1} > 40000\) (Either letter \(r\) or their \(r\) value) Allow ' = ' or ‘<’	M1
\(r^M > 1.6 \Rightarrow \log r^M > \log 1.6\) Allow ' = ' or ‘<’ (See below) OR (by change of base), \(\log_{1.03} 1.6 < M\ \Rightarrow\ \dfrac{\log 1.6}{\log 1.03} < M\)	M1
\((N - 1)\log 1.03 > \log 1.6\) (Correct bracketing required) (*) Accept work for part (c) seen in part (d)	A1 cso
	(3)

Notes

(c) 2^nd M: Requires \(\dfrac{40000}{25000}\) to be dealt with, and ‘two’ logs introduced.
With, say, \(N\) instead of \(N - 1\), this mark is still available.

Jumping straight from \(1.03^{N-1} > 1.6\) to \((N - 1)\log 1.03 > \log 1.6\) can score only M1 M0 A0.
(The intermediate step \(\log 1.03^{N-1} > \log 1.6\) must be seen).

Longer methods require correct log work throughout for 2^nd M, e.g.:\[\log(25000r^{N-1}) > \log 40000\ \Rightarrow\ \log 25000 + \log r^{N-1} > \log 40000\ \Rightarrow\]\[\log r^{N-1} > \log 40000 - \log 25000\ \Rightarrow\ \log r^{N-1} > \log 1.6\]

Mark scheme (d)

Scheme	Marks
(d) Attempt to evaluate \(\dfrac{\log 1.6}{\log 1.03} + 1\) {or \(25000(1.03)^{15}\) and \(25000(1.03)^{16}\) }	M1
\(N = 17\) (not 16.9 and not e.g. \(N \geqslant 17\) ) Allow ‘17^th year’ Accept work for part (d) seen in part (c)	A1
	(2)

Notes

(d) Correct answer with no working scores both marks.
Evaluating \(\log\left(\dfrac{1.6}{1.03}\right) + 1\) does not score the M mark.

Mark scheme (e)

Scheme	Marks
(e) Using formula \(\dfrac{a(1 - r^n)}{1 - r}\) with values of \(a\) and \(r\), and \(n\) = 9, 10 or 11	M1
\(\dfrac{25000(1 - 1.03^{10})}{1 - 1.03}\)	A1
287 000 (must be rounded to the nearest 1 000) Allow 287000.00	A1
	(3)
	10

Notes

(e) M1 can also be scored by a “year by year” method, with terms added.
(Allow the M mark if there is evidence of adding 9, 10 or 11 terms).
1^st A1 is scored if the 10 correct terms have been added (allow terms to be to the nearest 100).
To the nearest 100, these terms are:
25000, 25800, 26500, 27300, 28100, 29000, 29900, 30700, 31700, 32600

No working shown: Special case: 287 000 scores 1 mark, scored on ePEN as 1, 0, 0.
(Other answers with no working score no marks).