The sum of $n$ arithmetic means between $a$ and $b$ , is

\frac{n (a + b)}{2}

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n (a + b)

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\frac{(n + 1) (a + b)}{2}

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(n + 1) (a + b)

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Solution

The correct option is A $\frac{n (a + b)}{2}$
Let, a and b two given numbers.

Let,

$A_{1}, A_{2}, A_{3} . . . . . . . . . . . . . A_{n}$ be Arithmetic means between a and b.

$a, A_{1}, A_{2}, A_{3} . . . . . . . . . . . . . A_{n}, b$ are in A.P.

$a + A_{1} + A_{2} + A_{3} + . . . . . . A_{n} + b = (\frac{n + 2}{2}) (a + b)$ (first term is =a, last term=b, number of term=n+2)

$A_{1}, A_{2}, A_{3} . . . . . . . . . . . . . A_{n} = \frac{(n + 2) (a + b)}{2} - (a + b)$

$= (a + b) [\frac{n + 2}{2} - 1]$

$= (a + b) [\frac{n + 2 - 2}{2}]$

$= n (\frac{a + b}{2})$

$=$ n times the single A.M. Between a and b

Hence, this is the answer.

Suggest Corrections

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