# APGP Question #3 This is a question from a recent ACJC JC2 test.

A philanthropist started a donation matching programme to encourage more people to donate regularly to a particular charity.

(a) For a person who donates  more than the previous month, the philanthropist will donate 20 in the third month, and ten times more than John in the seventh month.

(ii) Find the total amount of money donated by John and the philanthropist in one year, leaving your answer to the nearest dollar.

(b) In a revised donation matching programme, if a donor makes a monthly donation of latex 0\% latex 10 \% latex 20 \% latex 30 \% latex 4^{th} 2c.

(ii) By expressing the total amount of money, the philanthropist will donate at the end of the month as a summation, and using the result , show that the total amount of money the philanthropist will donate at the end of the month is latex \frac{(n-1)n(n+1)}{30} c latex n^{th} latex a latex a latex a + b^2 latex ab latex a + 6b^2 latex ab^6 latex ab^2 = 20 latex ab^6 = 10(a+ 6b^2) latex b^2 = \frac{20}{a}latex \Rightarrow a(\frac{20}{a})^3 = 10[a + 6(\frac{20}{a})]latex a^3 + 120a – 800 = 0 latex a = 5.373610646 \approx 5.37 (3 SF)latex b = 1.929220642 \approx 1.93 (3 SF) latex = \frac{12}{2} [2(5.373610646) + (12-1)(1.929220642)^2] = 310.1282186 latex = \frac{5.373610646(1.929220642^{12} – 1)}{1.929220642 – 1} = 15366.20901 latex = 15366.20901 + 310.1282186 = 15676.33723 \approx 15676.34\$

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