Question

The fourth term of an AP is 0. Prove that its 25th term is triple its 11th term

Answer 1

let first term of Ap is a and common difference d
4th term =a+(4-1) d=a+3d=0
a=-3d
now
t25=a+(25-1) d=a+24d=-3d+24d=21d
t11=a+(11-1) d=a+10d=-3d+10d=7d
here it is clear that
t25=3 x t11

Answer 2

let first term of Ap is a and common difference d

4th term =a+(4-1) d=a+3d=0

a=-3d

now

t25=a+(25-1) d=a+24d=-3d+24d=21d

t11=a+(11-1) d=a+10d=-3d+10d=7d

here it is clear that

t25=3 x t11