If the mean deviation about the mean of the numbers 1,2,3,…,n, where n is odd, is n5(n+1), then n is equal to \_\_\_
Answer: 21
Solutions · 1
Official solution
Mean =2n+1.
\[\text{M.D.} = \frac{2\left(\frac{n-1}{2}+\frac{n-3}{2}+\cdots+0\right)}{n} = \frac{5(n+1)}{n}\]
\[(n-1)+(n-3)+\cdots+0 = 5(n+1)\]
The sum of that arithmetic series (with 2n+1 terms... using the recovered relation) gives 4n+1⋅(n−1)=5(n+1)⟹n−1=20⟹n=21.