If 9th term of an A.P. is zero, prove that its 29th term is double (twice) the 19th term.

Here, t₉ = 0

We have to prove that t₂₉ = 2t₁₉

We know that nth term of A.P. , t_n = a + (n – 1) d.

⇒ t₉ = a + (9 – 1) d = a + 8d

But t₉ = 0

⇒ a + 8d = 0

∴ a = -8d … (i)

Now, t₂₉ = a + (29 – 1) d

= a + 28d

= -8d + 28d [From (i)]

= 20d

∴ t₂₉ = 20d … (1)

Now, t₁₉ = a + (19 – 1) d

= a + 18d

= -8d + 18d [From (i)]

= 10d

∴ t₁₉ = 10d … (2)

Equating (1) and (2),

⇒ 20d = 10d

⇒ 20d = 10d (2)

∴ t₂₉ = 2(t₁₉)

Hence proved.