Question

If the 9^th term of an AP is zero, prove that its 29^th term is twice its 19^th term.

Answer 1

Let the first term, common difference and number of terms of an AP are a, d and n respectively.

Given that, 9^th term of an AP

i.e., T₉ = 0

∵ n^th term of an AP, T_n = a + (n – 1)d

⇒ a + (9 – 1)d = 0

⇒ a + 8d = 0

⇒ a = – 8 d   ........(i)

Now its 19^th term, T₁₉ = a + (19 – 1)d

= – 8d + 18d ......[From equation (i)]

= 10d   .........(ii)

And its 29^th term, T₂₉ = a + (29 – 1)d

= – 8d + 28d

= 20d

= 2 × (10d)

From equation (i)]

⇒ T₂₉ = 2 × T₁₉

Hence its 29^th term is twice its 19^th term.