Question

The 8th term of an arithmetic progression is zero.Prove that its 38th term is triple of its 18th term.

Answer 1

let a.p. is a,a+d,a+2d,a+3d...................,a+37d,........
according to question 8th term is zero
0 = a+(8-1)d
a = -7d --------------------(1)
18th term:
T18 = a+(18-1)d 
T18 = a+17d
put a = -7d from equation (1)
T18 = -7d+17d = 10d
38th term:
T38 = a+(38-1)d
T38 = a+37d
ptu a = -7d from equation (1)
T38 = -7d+37d = 30d
T38 = 3(10d) = 3T8
hence prove