Question

A set of consecutive natural numbers from 1 to $n$ are written and one number is removed from it
The arithmetic mean of the remaining numbers is found to be $\frac{71}{2}$, then the sum of all possible value(s) of the number removed from the set is $\lambda$, then find the value of $[\sqrt{\lambda }]$.
$[$ Note : $[x]$ denotes greatest integer less than or equal to $x$.

Answer 1

Answer: 8

$\frac{\frac{n(n+1)}{2}-n}{n-1}\le \frac{71}{2}\le \frac{\frac{n(n+1)}{2}-1}{n-1}$
$\frac{n^2-n}{2(n-1)}\le \frac{71}{2}\le \frac{n^2+n-2}{2(n-1)}$
$\frac{n}{2}\le \frac{71}{2}\le \frac{n+2}{2}$
Hence, $n$ can be 69,70 or 71 .
For $n=69,\frac{\frac{69(70)}{2}-x}{68}=\frac{71}{2}\Rightarrow x=1$
For $n=70,\frac{\frac{70(71)}{2}-x}{69}=\frac{71}{2}\Rightarrow$ we do not get $x\in N$
For $n=71$, we get $x=71$
Hence $\lambda =1+71=72$
$[\sqrt{72}]=8$