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Nov 2005 p1 q6
812
A small trading company made a profit of $250,000 in the year 2000. The company considered two different plans, plan A and plan B, for increasing its profits.
Under plan A, the annual profit would increase each year by 5% of its value in the preceding year. Find, for plan A,
(i) the profit for the year 2008,
(ii) the total profit for the 10 years 2000 to 2009 inclusive.
Under plan B, the annual profit would increase each year by a constant amount $D$.
(iii) Find the value of $D$ for which the total profit for the 10 years 2000 to 2009 inclusive would be the same for both plans.
Solution
(i) The profit follows a geometric progression with initial term \(a = 250,000\) and common ratio \(r = 1.05\). The year 2008 is the 9th term, so we calculate \(ar^8 = 250,000 \times 1.05^8 = 369,000\).
(ii) The total profit for 10 years is given by the sum of a geometric series: \(S_{10} = 250,000 \left( \frac{1.05^{10} - 1}{0.05} \right) = 3,140,000\).
(iii) Under plan B, the profit forms an arithmetic progression. The sum of the profits for 10 years is \(S_{10} = 5(500,000 + 9D)\). Setting this equal to the total from plan A, \(5(500,000 + 9D) = 3,140,000\), solving gives \(D = 14,300\).