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,

T₉ = 0   ...[∵ n^th term of an AP, T_n = a + (n – 1)d]

⇒ a + (9 – 1)d = 0

⇒ a + 8d = 0

⇒ a = – 8d   ...(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  ...[From equation (i)]

= 2 × (10d)   

= 20d

⇒ T₂₉ = 2 × T₁₉

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