Question

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

Answer 1

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

⇒ a = -3d

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, 25^th term is triple its 11^th term.