If x, y, z are in A.P.

x, y, z are in A.P.

$\therefore y=\frac{x+z}{2}$

Now, $A_{1}$ is the arithmetic mean of $\mathrm{x}$ and $\mathrm{y}$.

$A_{1}=\frac{x+y}{2}=\frac{x+\frac{x+z}{2}}{2}=\frac{3 x+z}{4}$

And, $A_{2}$ is the arithmetic mean of $y$ and $z$.

$A_{2}=\frac{y+z}{2}=\frac{\frac{x+z}{2}+z}{2}=\frac{3 z+x}{4}$

Let $A_{3}$ be the arithmetic mean of $A_{1}$ and $A_{2}$.

$A_{3}=\frac{A_{1}+A_{2}}{2}$

$=\frac{\frac{3 x+z}{4}+\frac{3 z+x}{4}}{2}$

$=\frac{4 x+4 z}{8}$

$=\frac{x+z}{2}$

$=y$

Hence proved.