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

Given:

$a_{9}=0$

$\Rightarrow a+(9-1) d=0 \quad\left[a_{n}=a+(n-1) d\right]$

$\Rightarrow a+8 d=0$

$\Rightarrow a=-8 d \ldots(\mathrm{i})$

To prove.

$a_{29}=2 a_{19}$

Proof:

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

$=a+28 d$

$=-8 d+28 d \quad($ From $(\mathrm{i}))$

$=20 d$

RHS : $2 a_{19}=2[a+(19-1) d]$

$=2(a+18 d)$

$=2 a+36 d$

$=2(-8 d)+36 d \quad($ From $(\mathrm{i}))$

$=-16 d+36 d$

$=20 d$

LHS = RHS

Hence, proved.