Question

If x,y, z are in A.P. and $A_1$ is the A.M. of x and y and $A_2$ is the A.M. of y and z, then prove that the A.M. of $A_1$ and $A_2$ is y.

Answer 1

x, y, z are in A.P.

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

Now, $A_1$ is the arithmetic mean of x and y.

$A_1=\frac{x+y}{2}=\frac{x+\frac{x+z}{2}}{2}=\frac{3x+z}{4}$

And, $A_2$ is the arithmetic mean of y and z.

$A_2=\frac{y+z}{2}=\frac{\frac{x+z}{2}+z}{2}=\frac{3z+x}{4}$

Let $A_3$ be the arithmetic mean of $A_1$ and $A_2$.

$A_3=\frac{A_1+A_2}{2}$

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

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

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

$=y$

Hence proved.