If the 9th term of an AP is zero,

Let the first term, common difference and number of terms of an AP are a, d and n respectively.

Given that, $\quad$ 9th term of an AP, $T_{9}=0 \quad\left[\because n\right.$th term of an AP, $\left.T_{n}=a+(n-1) d\right]$

$\Rightarrow \quad a+(9-1) d=0$

$\Rightarrow \quad a+8 d=0 \Rightarrow a=-8 d$ $\ldots$ (i)

Now, its 19th term, $T_{19}=a+(19-1) d$ 

$=-8 d+18 d \quad$ [from Eq. (i)]

$=10 d$ ... (ii)

and its 29th term, $T_{29}=a+(29-1) d$

$=-8 d+28 d$

$=-8 d+28 d \quad$ [from Eq. (i)]

$=20 d=2 \times(10 d)$

$\Rightarrow$ $T_{29}=2 \times T_{19}$

Hence, its 29th term is twice its 19th term.