$\begin{array}{1 1}(A)\;\large\frac{n}{4}\\(B)\;(n+1)\\(C)\;0\\(D)\;\large\frac{n-1}{12}\end{array} $

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Mean of first n natural numbers where n is an even number is $= \large\frac{1}{2} \bigg[ \frac{n}{2} +\frac{n+1}{2} \bigg]$

$\qquad= \large\frac{2n+1}{4}$

mean deviation $=\large\frac{\sum |x_i -x_n|}{n}$

$\qquad= \large\frac{1}{4n} $$ [1+3+5+9+.......+nth\; odd number]$

now sum of odd numbers $=n^2 $

$\qquad=\large\frac{1}{4n}$$n^2$

$\qquad= \large\frac{n}{4}$

Hence A is the correct answer.

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eg. 1, 2, 3, 4, 5, 6 are first 6(even) natural numbers.

mean= (1+2+3+4+5+6)/6 = 21/6 = 3.5 ----------- (1)

now (2*6 + 1)/4 = 13/4 = 3.25 ----------------------- (2)

clearly 1&2 are not equal