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.