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The weighted A.M of first $n$ natural numbers whose weights are equal is

$\begin {array} {1 1} (A)\;\large\frac{n+1}{2} & \quad (B)\;\large\frac{2n+1}{2} \\ (C)\;\large\frac{2n+1}{3} & \quad (D)\;\large\frac{(2n+1)(n+1)}{6} \end {array}$


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1 Answer

+1 vote
Let weight of each is $w$ , then weight
A.M = $ \large\frac{w.1+w.2+.....+w.n}{w+w+w.....+w}$
$ = \large\frac{1+2+....+n}{n}$
$ = \large\frac{n(n+1)}{2.n}$
$ = \large\frac{n+1}{2}$
Ans : (A)
answered Jan 25, 2014 by thanvigandhi_1

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