If $a_{1},a_{2},a_{3},\ldots,a_{n}$ are the $n$ arithmetic means between $a$ and $b$, then $2\sum_{i = 1}^{n}a_{i}$ equals

Question

Dr. Abubacker · Accepted Answer

(b)
Since, $a_{1},a_{2},\ldots,a_{n}$ are $n$ AM’s between $a$ and $b$, then
$a_{i} = a + i\ \frac{(b - a)}{n + 1}$
$	herefore 2\sum_{i = 1}^{n}a_{i} = 2\left\lbrack \sum_{}^{}{a + \frac{(b - a)}{n + 1}}\ \sum_{}^{}i ightbrack$
$= 2\left\lbrack na + \frac{(b - a)}{n + 1}.\ n\ \frac{(n + 1)}{2} ightbrack$
$= n(a + b)$

Q. If $a_{1},a_{2},a_{3},\ldots,a_{n}$ are the $n$ arithmetic means between $a$ and $b$, then $2\sum_{i = 1}^{n}a_{i}$ equals

Explanation

Reviewed by Dr. Abubacker

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