Given,

9^th term of an A.P is 0

⇒ a₉ = 0

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

⇒ a₉ = a + 8d

According to question:

a₉ = 0

⇒ a + 8d = 0

⇒ a = -8d

When n = 19:

∴ a₁₉ = a + (19 – 1)d

⇒ a₁₉ = a + 18d

⇒ a₁₉ = -8d + 18d

⇒ a₁₉ = 10d

When n = 29:

∴ a₂₉ = a + (29 – 1)d

⇒ a₂₉ = a + 28d

{∵ a = -8d}

⇒ a₂₉ = -8d + 28d

⇒ a₂₉ = 20d

⇒ a₂₉ = 2×10d

{∵ a₁₉ = 10d}

⇒ a₂₉ = 2a₁₉

Hence Proved