From a set of positive consecutive integers starting with $1$, one number is erased and the AM of the remaining numbers is $\frac{602}{17}$ A set of positive consecutive integers starting with one is written on a blackboard. One number is erased and the AM of the remaining numbers is $\frac{602}{17}$. The erased number is 1. 6 2. 7 3. 8 4. 9 Initially, $$\text{AM}=\frac{n+1}{2}$$ After removing the number $x$, $$\text{AM}=\frac{\frac{n(n+1)}{2}-x}{n-1}=\frac{602}{17}$$ How can I solve the above equation?

Finally I came up with a solution.

$$\frac{\frac{n(n+1)}{2}-n}{n-1}\le \frac{602}{17}\le\frac{\frac{n(n+1)}{2}-1}{n-1}$$ $$\frac{n}{2}\le\frac{602}{17}\le\frac{n+2}{2}$$ $$n\le70+\frac{14}{17}\le n+2$$ $$n=69 \text{ or } 70$$

Substituting $n=69$ the actual equation, $x=7$

$n=70$ is not possible since I got a non integral $x$