Question

Show that the sum of the arithmetic means, between two given quantities, equidistant from the beginning and end is constant.

Answer 1

To show sum of $AM$, between $2$ given densities, equal from beginning & end is constant

$\Rightarrow$ Let $K,L$ be a constant from beginning & end & $k,l,m$ be variables for any $2$ no: & equidistant also

$\therefore k=K+m$

$l=L-m$

So Arithmetic mean $\Rightarrow \frac{k+l}{2}=\frac{K+m+L-m}{2}=\frac{K+L}{2}$

It is a constant.

Hence proved.