The Chief Examiner reported that this question was very popular among the candidates and they performed very well in it. Candidates were reportedly able to show that if the sum to infinity was 25, then a = 25 – 25r, where a = first term and r = constant ratio. Also, since the sum of the first two terms was 16, then, a + ar = 16 i.e. a(1+ r) = 16. Solving these two equations simultaneously gave the quadratic equation 25(1 - r2) = 16. Solving this equation gave r = . Substituting for r in the equation a = 25 – 25r gave a = 10. The fifth term was obtained using the formula T5 = ar4 = 10. The sum of the first four terms = 10() = 10() = 25 ´ = = 21.