The 4^th term of an A. P. is zero. Prove that the 25^th term of the A.P. is three times its 11^th term. (CBSE 2016)

Let a and d be first term and common difference of AP respectively.

Then we know that nth term of an AP is

a_n = a + (n - 1)d

Now given a₄ = 0

a + 3d = 0

a = - 3d [eqn 1]

and

a₂₅ = a + 24d

= - 3d + 24d [using eqn 1]

= 21d

a₁₁ = a + 10d

= - 3d + 10d [using eqn 2]

= 7d

So we have, a₂₅ = 3a₁₁

i.e. 25^th term of the AP is three times its 11^th term