Question

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

Answer 1

Mean$=\frac{n\frac{(n+1)}{2}}{n}=\frac{n+1}{2}$
$\text{M.D.}=\frac{2(\frac{n-1}{2}+\frac{n-3}{2}+\frac{n-5}{2}+\dots 0)}{n}=\frac{5(n+1)}{n}$
$\Rightarrow \left(\right(n-1)+(n-3)+(n-5)+\dots 0)=5(n+1)$
$\Rightarrow \left(\frac{n+1}{4}\right)\cdot (n-1)=5(n+1)$
$So,n=21$