It was reported that this question was attempted by majority of the candidates and their performance was commendable.
In part (a), candidates were able to substitute correctly in the formula. Tn = arn-1, (where a = first term, r = constant ratio, n= required term and Tn = value of the nth term), to obtain two equations, ar5 = 54 and ar2 = 2. To eliminate a the first equation is divided by the second one, i.e. ar5 ar2 = 54/2 ; r³ = 27. Thus, r = 3. This value for r is substituted into any of the equation to get a .i.e. To get a, a(3)2 = 2 ; a =2/9 . The sum of the first 10 terms was gotten by substituting correctly in the formula
Sn = a (r10-1) = 2/9 (310-1)
r – 1 2 = 6561 to the nearest whole number.
In part (b) candidates’ performance was reported to be good though not as good as it was in part (a). Candidates were expected to obtain the ratio of the coefficients of x4 and x3 in the expansion of (1+2x)n and equate it to 3:1 thus:
Co-efficient of x4 = n(n-1)(n-2)(n-3)24, coefficient of x3 = n(n-1)(n-2)23 .
4! 3!
Hence, n(n-1)(n-2)(n-3) 24 ¸ n(n-1)(n-2)23 = 3 i.e. n(n-1)(n-2)(n-3)243!= 3
4! 3! 1 n(n-1)(n-2)234! 1
Simplifying the expression gave n = 9.