a = 10, d = – 3

(ii) a = 10, d = – 3 (1 mark)

Sol. a = 10, d = – 3

Here, t1 = a = 10

t2 = t1 + d = 10 + (– 3) = 10 – 3 = 7

t3 = t2 + d = 7 + (– 3) = 7 – 3 = 4

t4 = t3 + d = 4 + (– 3) = 4 – 3 = 1

t5 = t4 + d = 1 + (– 3) = 1 – 3 = – 2

∴ The first five terms of the A.P. are 10, 7, 4, 1 and – 2.