Question

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

Answer 1

4th term of an A.P. = a₄ = 0

∴ a + (4 – 1) d = 0

∴ a + 3d = 0

∴ a = –3d   ...(1)

25^th term of an A.P. = a₂₅

= a + (25 – 1)d

= –3d + 24d   ...[From (1)]

= 21d.

3 times 11^th term of an A.P. = 3a₁₁

= 3[a + (11 – 1)d]

= 3[a + 10d]

= 3[–3d + 10d]

= 3 × 7d

= 21d

∴ a₂₅ = 3a₁₁

i.e., the 25^th term of the A.P. is three times its 11^th term.

Answer 2

Let the A.P. have first term a and common difference d.

The nth term is a_n = a + (n − 1)d.

a₄ ​= a + 3d = 0 ⇒ a = −3d.

a_25​ = a + 24d = (−3d) + 24d = 21d,

a₁₁ ​= a + 10d = (−3d) + 10d = 7d.

a₂₅ ​= 21d = 3 ⋅ 7d = 3a_11​.