If the 4th, 10th and 16th terms of a G.P. are x, y and z, respectively. Prove that x, y, z are in G.P.

Let a be the first term and r be the common ratio of the G.P.

According to the given condition,

$a_{4}=a r^{3}=x \ldots(1)$

$a_{10}=a r^{9}=y \ldots(2)$

$a_{16}=a r^{15}=z \ldots$ (3)

Dividing $(2)$ by $(1)$, we obtain

$\frac{y}{x}=\frac{a r^{9}}{a r^{3}} \Rightarrow \frac{y}{x}=r^{6}$

Dividing (3) by (2), we obtain

$\frac{z}{y}=\frac{a r^{15}}{a r^{9}} \Rightarrow \frac{z}{y}=r^{6}$

$\therefore \frac{y}{x}=\frac{z}{y}$

Thus, $x, y, z$ are in G. P.