If A is an arithmetic mean between two numbers and S is the sum of n arithmetic means between the same number, then

Let the numbers be $a$ and $b.$

Then,  $a_{1}A,b\text{ are in }AP$ respectively.

$\therefore A=\frac{a+b}{2}$

Let $A_1,A_2,A_3,\cdots ,A_n$ are $n$ arithmetic means between $a$ and $b.$

Then, $a_1,A_1,A_2,\cdots ,A_n,b$ are in $AP$.

$\Rightarrow b=a+(n+1)d$ where d is the common difference.

$\begin{array}{r}d=\frac{b-a}{n+1}\\ A_1=a+d=a+\frac{b-a}{n+1}\\ =\frac{na+a+ba}{n+1}=\frac{na+b}{n+1}\end{array}$

Since. S is the sum of n arithmetic means. 

$\begin{array}{l}\therefore S=\frac{n}{2}\left[2\frac{(na+b)}{n+1}+(n-1)\frac{b-a}{n+1}\right]\\ =\frac{n}{2(n+1)}[2na+2b+(n-1\left)\right(b-a\left)\right]\\ =\frac{n}{2(n+1)}[an+nb+b+a]\\ =\frac{n}{2(n+1)}\left[n\right(a+b)+(a+b\left)\right]\\ =\frac{n}{2(n+1)}\left[\right(n+1\left)\right(a+b\left)\right]\\ =\frac{n(a+b)}{2}\\ =n\cdot A\end{array}$