Let the first term of the given geometric progression (GP) be $a$ and the common ratio be $r$. The $n^{th}$ term of a GP is given by the formula:

$$t_n = ar^{n-1}$$

We are given that the $4^{th}$, $10^{th}$, and $16^{th}$ terms of this GP are $x$, $y$, and $z$ respectively.

Using the formula for the $n^{th}$ term of a GP, we can express $x$, $y$, and $z$ in terms of $a$ and $r$:

$$x = ar^{4-1} = ar^3$$ $$y = ar^{10-1} = ar^9$$ $$z = ar^{16-1} = ar^{15}$$

To prove that $x$, $y$, and $z$ are in a geometric progression, we must show that the ratio of consecutive terms is constant. That is, we need to prove that:

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

Let us calculate the ratio $\frac{y}{x}$. Substituting the expressions for $y$ and $x$, we get:

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

Now, let us calculate the ratio $\frac{z}{y}$. Substituting the expressions for $z$ and $y$, we get:

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

Since $\frac{y}{x} = \frac{z}{y} = r^6$, the terms $x$, $y$, and $z$ have a constant common ratio.

∴ $x$, $y$, $z$ are in GP.