The $9^{th}$ term of an $A.P.$ is $0.$ Show that the $29^{th}$ term of the $A.P.$ is two times its $19^{th}$ term.

(N/A) Let the first term of the $A.P.$ be $a$ and the common difference be $d.$
The $n^{th}$ term of an $A.P.$ is given by $a_n = a + (n - 1)d.$
Given that the $9^{th}$ term is $0$,we have:
$a_9 = a + (9 - 1)d = 0$
$a + 8d = 0$
$a = -8d$ --- $(1)$
Now,we need to find the $29^{th}$ term:
$a_{29} = a + (29 - 1)d = a + 28d$
Substituting $a = -8d$ from $(1)$:
$a_{29} = -8d + 28d = 20d$ --- $(2)$
Next,we find the $19^{th}$ term:
$a_{19} = a + (19 - 1)d = a + 18d$
Substituting $a = -8d$ from $(1)$:
$a_{19} = -8d + 18d = 10d$ --- $(3)$
Comparing $(2)$ and $(3)$,we see that $a_{29} = 20d$ and $2 \times a_{19} = 2 \times 10d = 20d.$
Therefore,$a_{29} = 2 \times a_{19}.$ Hence,the $29^{th}$ term is two times its $19^{th}$ term.