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

$\frac{\frac{ n ( n +1)}{2}- n }{ n -1} \leq \frac{71}{2} \leq \frac{\frac{ n ( n +1)}{2}-1}{ n -1}$
$\frac{ n ^2- n }{2( n -1)} \leq \frac{71}{2} \leq \frac{ n ^2+ n -2}{2( n -1)} $
$\frac{ n }{2} \leq \frac{71}{2} \leq \frac{ n +2}{2}$
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 $\lambda=1+71=72$
$[\sqrt{72}]=8$