If the 9th term of an A.P. is zero then show that the 29th term is twice the 19th term.

By using n^th term of an A.P. formula
t_n = a + (n – 1) d
where n = number of terms
a = first term
d = common difference
t_n = n^th terms
Given: t₉ = 0
⇒ t₉ = a + (9 – 1) d
⇒ 0 = a + 8d
⇒ a = – 8d
To Show: t₂₉ = 2× t₁₉
Now,
⇒ t₂₉ = a + (29 – 1) d
⇒ t₂₉ = a + 28d
⇒ t₂₉ = – 8d + 28d = 20 d (since, a = – 8d)
⇒ t₂₉ = 20 d
⇒ t₂₉ = 2 × 10 d …. (1)
Also,
⇒ t₁₉ = a + (19 – 1) d
⇒ t₁₉ = a + 18d
⇒ t₁₉ = – 8d + 18d = 10 d (since, a = – 8d)
⇒ t₁₉ = 10 d …. (2)
From equation (1) and equtaion (2) we get,
t₂₉ = 2 × t₁₉