If the  ${19}^{th}$   term of a non-zero A.P. is zero, then its  ( ${49}^{th}$ term) : ( ${29}^{th}$ term) is

We know that the general term of an arithmetic progression having first term $a$ common difference $d$ is $a_n=a+\left(n-1\right)d$

Given that 19th term is zero, so that $a+\left(19-1\right)d=0$

It implies that $a+18d=0$

We want to find the ratio of 49th term and 29th terms is $T_{49}:T_{29}$

Hence, 

$\begin{array}{rcl}\frac{T_{49}}{T_{29}}& =& \frac{a+48d}{a+28d}\\ & =& \frac{-18d+48d}{-18d+28d}\\ & =& \frac{30d}{10d}\\ & =& \frac{3}{1}\end{array}$

Therefore, the ratio of 49th term and 29th term is $\boxed{\mathbf{3}\mathbf{:}\mathbf{1}}$