If A is an arithmetic mean between two number | vedclass

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(A) Let the two numbers be $a$ and $b$.The arithmetic mean $A$ between $a$ and $b$ is given by $A = \frac{a+b}{2}$.Let $A_1, A_2, \dots, A_n$ be the $

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$S = nA$

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(A) Let the two numbers be $a$ and $b$.The arithmetic mean $A$ between $a$ and $b$ is given by $A = \frac{a+b}{2}$.Let $A_1, A_2, \dots, A_n$ be the $n$ arithmetic means between $a$ and $b$. Then $a, A_1, A_2, \dots, A_n, b$ form an arithmetic progression $(A.P.)$ with $n+2$ terms.The common difference $d$ is given by $d = \frac{b-a}{n+1}$.The sum $S$ of $n$ arithmetic means is $S = A_1 + A_2 + \dots + A_n$.Since $A_1, A_2, \dots, A_n$ are in $A.P.$,their sum is $S = \frac{n}{2}(A_1 + A_n)$.Here,$A_1 = a + d = a + \frac{b-a}{n+1} = \frac{an + a + b - a}{n+1} = \frac{an + b}{n+1}$ and $A_n = b - d = b - \frac{b-a}{n+1} = \frac{bn + b - b + a}{n+1} = \frac{bn + a}{n+1}$.Thus,$S = \frac{n}{2} \left( \frac{an + b + bn + a}{n+1} \right) = \frac{n}{2} \left( \frac{(a+b)(n+1)}{n+1} \right) = \frac{n(a+b)}{2}$.Since $A = \frac{a+b}{2}$,we have $S = nA$.

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

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