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

Let the GP has first term A and common ratio R.

$$x=t_4=A\cdot R^3$$

$$y = t_{10} = A \cdot R^9$$

$$z = t_{16} = A \cdot R^{15}$$

Check the ratio:

$$\frac{y}{x} = \frac{(A \cdot R^9)}{(A \cdot R^3)} = R^6$$

$$\frac{z}{y} = \frac{(A \cdot R^{15})}{(A \cdot R^9)} = R^6$$

Since $\frac{y}{x} = \frac{z}{y} = \frac{R}{6}$, the three numbers $x$, $y$, $z$ have a common ratio $r = 6$.
Therefore, $x$, $y$, and $z$ are in GP.