Average of a sequence

The average is calculated by $\frac{1 + 2 + 3 + ... + 2x + ... + n}{n} = \frac{1 + 2 + ... + n + x}{n} = \frac{ \frac{n(n+1)}{2} + x}{n} = \frac{n+1}{2} + \frac{x}{n}$ . And the problem tells us that $\frac{n+1}{2} + \frac{x}{n} = 51$ . With a little algebra we can re-arrange this into $x = \frac{101n - n^2}{2}$ . Unfortunately we really have no idea what $n$ has to be, but there are two properties we have yet to apply. First, $x$ has to be a positive integer so we have the inequality $101n - n^2 > 0 \implies 101n > n^2 \implies 101 > n$ . The other property is that $x \leq n$ so we have $\frac{101n - n^2}{2} \leq n \implies 101n - n^2 \leq 2n \implies 99n \leq n^2 \implies 99 \leq n$ . So what we have found is that $101 > n \geq99 \implies n=100,99$ . And using the formula we have, $x = 50$ when $n=100$ and $x=99$ when $n=99$ . Thus the maximum value is $n - x = 50$ .