The 4th term of an AP is zero. Prove that its 25 term is triple is 11th term.

In the given AP, let the first term be a and the common difference be d.
Then, Tn = a + (n - 1)d ​
Now, T4 = a + (4 - 1)d 
⇒ a + 3d  = 0        ...(1)

Again, T11 = a + (11 - 1)d  = a + 10d
=​ -3d + 10 d = 7d             [ Using (1)]

Also, T25 = a + (25 - 1)d =​ a + 24d = -3d + 24d = 21d             [ Using (1)]​
    i.e., T25 = 3 ⨯ 7d = (3 ⨯ T11)
​Hence, 25th term is triple its 11th term.