If the $9^{\text{th}}$ term of an $AP$ is zero,prove that its $29^{\text{th}}$ term is twice its $19^{\text{th}}$ term.

(N/A) Let the first term be $a$ and the common difference be $d$ for an $AP$.
The $n^{\text{th}}$ term of an $AP$ is given by the formula $T_n = a + (n - 1)d$.
Given that the $9^{\text{th}}$ term is zero,we have $T_9 = 0$.
$a + (9 - 1)d = 0$
$a + 8d = 0$
$a = -8d$ ....$(i)$
Now,calculate the $19^{\text{th}}$ term $(T_{19})$:
$T_{19} = a + (19 - 1)d$
$T_{19} = a + 18d$
Substituting $a = -8d$ from equation $(i)$:
$T_{19} = -8d + 18d = 10d$ ....$(ii)$
Next,calculate the $29^{\text{th}}$ term $(T_{29})$:
$T_{29} = a + (29 - 1)d$
$T_{29} = a + 28d$
Substituting $a = -8d$ from equation $(i)$:
$T_{29} = -8d + 28d = 20d$ ....$(iii)$
From equations $(ii)$ and $(iii)$:
$T_{29} = 20d = 2 \times (10d) = 2 \times T_{19}$.
Thus,the $29^{\text{th}}$ term is twice its $19^{\text{th}}$ term.