If the $4^{\text{th}}$,$10^{\text{th}}$,and $16^{\text{th}}$ terms of a $G.P.$ are $x, y$,and $z$ respectively,prove that $x, y, z$ are in $G.P.$

(N/A) Let $a$ be the first term and $r$ be the common ratio of the $G.P.$
According to the given condition:
$a_{4} = ar^{3} = x$ $(1)$
$a_{10} = ar^{9} = y$ $(2)$
$a_{16} = ar^{15} = z$ $(3)$
Dividing $(2)$ by $(1)$,we obtain:
$\frac{y}{x} = \frac{ar^{9}}{ar^{3}} = r^{6}$
Dividing $(3)$ by $(2)$,we obtain:
$\frac{z}{y} = \frac{ar^{15}}{ar^{9}} = r^{6}$
Since $\frac{y}{x} = \frac{z}{y} = r^{6}$,the ratio between consecutive terms is constant.
Therefore,$x, y, z$ are in $G.P.$