The 4th term of an A.P. is zero. Prove that the 25th term of the A.P. is three times its 11th term.

Let a be the first term and d be the common difference of an A.P. having n terms.
We know that nth term formula as, an = a + (n-1)d
As per the question, a4 = a + 3d = 0
a + 3d = 0
a = -3d
Need to prove a25 = a + 24d = 3(a + 10d)
a25 = -3d + 24d
a25 = 21d ——— (1)
a11 = a + 10d
a11 = -3d + 10d
a11 = 7d —— (2)
From equation (1) and (2), we get
a25 = 3a11
Hence proved.