If 9th term of an A.P is zero,

In the given problem, the 9th term of an A.P. is zero.

Here, let us take the first term of the A.P as a and the common difference as d

So, as we know,

$a_{n}=a+(n-1) d$

We get,

$a_{9}=a+(9-1) d$

$0=a+8 d$

$a=-8 d$ .......(1)

Now, we need to prove that $29^{\text {th }}$ term is double of $19^{\text {th }}$ term. So, let us first find the two terms.

For $19^{\text {th }}$ term $(n=19)$,

$a_{19}=a+(19-1) d$

$=-8 d+18 d \quad($ Using 1 $)$

$=10 \mathrm{~d}$

For 29th term (n = 29),

$a_{29}=a+(29-1) d($ Using 1)

$=-8 d+28 d$

$=20 d$

$=2 \times 10 d$

$=2 \times a_{19}$

Therefore, for the given A.P. the 29th term is double of the 19th term.

Hence proved.