Question

If 9^th term of an A.P. is Zero, prove that its 29^th term is double the 19^th term.

Answer 1

Given:

9^th term of an A.P is 0

So, a₉ = 0

We need to prove: a₂₉ = 2a₁₉

We know, a_n = a + (n – 1) d [where a is first term or a₁ and d is common difference and n is any natural number]

When n = 9:

a₉ = a + (9 – 1)d

= a + 8d

According to question:

a₉ = 0

a + 8d = 0

a = -8d

When n = 19:

a₁₉ = a + (19 – 1)d

= a + 18d

= -8d + 18d

= 10d

When n = 29:

a₂₉ = a + (29 – 1)d

= a + 28d

= -8d + 28d [Since, a = -8d]

= 20d

= 2×10d

a₂₉ = 2a₁₉ [Since, a₁₉ = 10d]

Hence Proved.