Question:
If n A.M.s are inserted between two numbers, prove that the sum of the means equidistant from the beginning and the end is constant.
Solution:
Let $A_{1}, A_{2} \ldots \ldots A_{n}$ be $n$ A.M.s between two numbers $a$ and $b$.
Then, $a, A_{1}, A_{2} \ldots \ldots A_{n}, b$ are in A.P. with common difference, $d=\frac{b-a}{n+1}$.
$\therefore A_{1}+A_{2}+\ldots \ldots+A_{n}=\frac{n}{2}\left[A_{1}+A_{n}\right]$
$=\frac{n}{2}\left[A_{1}-d+A_{n}+d\right]$
$=\frac{n}{2}[a+b]$
$=n \times\left[\frac{a+b}{2}\right]$
$=$ A.M. between $a$ and $b$, which is constant.