If the mean deviation about the mean of the n | vedclass

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(D) The mean of the first $n$ natural numbers is $\bar{x} = \frac{n+1}{2}$.The mean deviation about the mean is given by $\frac{1}{n} \sum_{i=1}^{n} |

Vedclass · Accepted Answer

$21$

Vedclass · Answer

(D) The mean of the first $n$ natural numbers is $\bar{x} = \frac{n+1}{2}$.The mean deviation about the mean is given by $\frac{1}{n} \sum_{i=1}^{n} |x_i - \bar{x}|$.For odd $n$,the mean deviation about the mean of the first $n$ natural numbers is $\frac{n^2-1}{4n}$.Given that the mean deviation is $\frac{5(n+1)}{n}$,we set up the equation:$\frac{n^2-1}{4n} = \frac{5(n+1)}{n}$.Since $n^2-1 = (n-1)(n+1)$,we have:$\frac{(n-1)(n+1)}{4n} = \frac{5(n+1)}{n}$.Dividing both sides by $\frac{n+1}{n}$ (since $n 
eq -1$):$\frac{n-1}{4} = 5$.$n-1 = 20$.$n = 21$.

$x$	$6$	$12$	$18$	$24$	$30$	$36$	$42$
$f$	$4$	$7$	$9$	$18$	$15$	$10$	$5$

If the mean deviation about the mean of the numbers $1, 2, 3, \ldots, n$,where $n$ is odd,is $\frac{5(n+1)}{n}$,then $n$ is equal to

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