Question

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

Answer 1

Let a be the first term and d be the common difference of an AP.

a₄ = a + (n- 1)d

= a + (4 – 1)d

= a + 3d

Since 4th term of an AP is zero.

a + 3d = 0

or a = -3d ….(1)

Similarly,

a₂₅ = a + 24d = -3d + 24d = 21d …(2)

a₁₁ = a + 10d = -3d + 10d = 7d ….(3)

From (2) and (3), we have

a₂₅ = 3 x a₁₁

Hence proved.