If n A.M.s are inserted between two numbers,

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.

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