C2 June 2007 Q8

EdexcelOld spec9 marksSequences & Series

8. A trading company made a profit of £50 000 in 2006 (Year 1).

A model for future trading predicts that profits will increase year by year in a geometric sequence with common ratio \(r,\ r \gt 1\).

The model therefore predicts that in 2007 (Year 2) a profit of £50 000\(r\) will be made.

(a) Write down an expression for the predicted profit in Year \(n\). (1)

The model predicts that in Year \(n\), the profit made will exceed £200 000.

(b) Show that \(n \gt \dfrac{\log 4}{\log r} + 1\). (3)

Using the model with \(r = 1.09\),

(d) find the total of the profits that will be made by the company over the 10 years from 2006 to 2015 inclusive, giving your answer to the nearest £10 000. (3)

Scheme	Marks
\(50\,000r^{n-1}\) (or equiv.) (Allow \(ar^{n-1}\) if \(50\,000r^{n-1}\) is seen in (b))	B1
	(1)

Mark scheme (b)

Scheme	Marks
\(50\,000r^{n-1} \gt 200\,000\) (Using answer to (a), which must include \(r\) and \(n\), and 200 000) (Allow equals sign or the wrong inequality sign) (Condone ‘slips’ such as omitting a zero)	M1
\(r^{n-1} \gt 4 \quad\Rightarrow\quad (n - 1)\log r \gt \log 4\) (Introducing logs and dealing correctly with the power) (Allow equals sign or the wrong inequality sign)	M1
\(n \gt \dfrac{\log 4}{\log r} + 1\) (*)	A1cso
	(3)

Notes

Incorrect inequality sign at any stage loses the A mark.
Condone missing brackets if otherwise correct, e.g \(n - 1\log r \gt \log 4\).

A common mistake: \(50\,000r^{n-1} \gt 200\,000\) M1
\(\phantom{\text{A common mistake: }}(n - 1)\log 50\,000r \gt \log 200\,000\) M0
(‘Recovery’ from here is not possible).

Mark scheme (c)

Scheme	Marks
\(r = 1.09\): \(\;n \gt \dfrac{\log 4}{\log 1.09} + 1\) or \(n - 1 \gt \dfrac{\log 4}{\log 1.09}\) \((n \gt 17.086...)\) (Allow equality)	M1
Year 18 or 2023 (If one of these is correct, ignore the other)	A1
	(2)

Notes

Correct answer with no working scores full marks.
Year 17 (or 2022) with no working scores M1 A0.
Treat other methods (e.g. “year by year” calculation) as if there is no working.

Mark scheme (d)

Scheme	Marks
\(S_n = \dfrac{a(1 - r^n)}{1 - r} = \dfrac{50000(1 - 1.09^{10})}{1 - 1.09}\)	M1 A1
£760 000 (Must be this answer… nearest £10000)	A1
	(3)
	9

Notes

M1: Use of the correct formula with \(a = 50000\), 5000 or 500000, and \(n = 9\), 10, 11 or 15.

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, 11 or 15 terms).
1^st A1 is scored if 10 correct terms have been added (allow “nearest £100”).
(50000, 54500, 59405, 64751, 70579, 76931, 83855, 91402, 99628, 108595)

No working shown: Special case: 760 000 scores 1 mark, scored as 1, 0, 0.
(Other answers with no working score no marks).