(a) The nth term ofa sequence is given by T; = 4Tn_l-3. If twice the third
term is five times the second term, find the first three terms of the
sequence.
~) Given ilia{ J -! 1] GJ {2~' find the values of constants k, m and r.

Candidates' expected responses were as follows:
In part (a), T, = 4 Tn-l -3, Hence, T2 = 4T1 -3 and T3 = 4 T2 - 3 = 4 (4T1 -3) - 3
= 16Tt - 15. Since 5T2 = 2T3, then 5(4Tj -3) = 2(16Tt -15) which after a little

computation gave T, = ~ or l~. Substituting ~ for T, in the formula Tn = 4Tn-1 - 3
4 4 4
gave the first three terms as 5/4, 2, 5.
In part (b), candidates were expected to obtain three equations» K - r = 2, 3m - r = 3 and 4m + 6r = 26. Solving the second and third equations
simultaneously gave m =2, and r = 3. Hence K = 5. The chief examiner commended the candidates' performance in this question as
majority of their responses were very good.