<s>[INST] Two consecutive numbers are removed from the progression 1, 2, 3..n. The arithmetic mean of the remaining numbers is 105/4. What is the value of n? Two consecutive numbers are removed from the progression 1, 2, 3..n. The arithmetic mean of the remaining numbers is 105/4. What is the smallest value of n?
?What are the two numbers removed?
Now many answers are given but they are given in a hit and trial method.
Is there any other method to solve this problem?
 [/INST] $\sum _{k=1}^n k= \dfrac{1}{2} n (n+1)$
Subtract $x+(x+1)$. The sum is then $\dfrac{1}{2} n (n+1) -2x-1$
The mean is then $\dfrac{\frac{1}{2} n (n+1) -2x-1}{n-2}=\dfrac{105}{4}$
$x=\dfrac{1}{8} \left(2 n^2-103 n+206\right)$ and $0<x<n$
$x>0 \to 2 n^2-103 n+206>0$ which means $n\ge 50$
And $x<n \to \dfrac{1}{8} \left(2 n^2-103 n+206\right)<n $ which means $2\le n\le 53$
so we have $50\le n\le 53$ but the only value that gives an integer $x$ is $n=50$
Finally the solution is $x=7$ when $n=50$
</s>