Question

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

Answer 1

In the given AP, let the first 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 + 10d

= 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.