7.1. Natural numbers from 1 to some $n$ are written in a row. When one of the numbers was removed, it turned out that the arithmetic mean of the remaining numbers is $40 \frac{3}{4}$. Find the number that was removed.
Answer: 61.

Solution. The sum of numbers from 1 to $n$ is $S_{n}=\frac{n(n+1)}{2}$. Let the number removed be $m$, where $1 \leqslant m \leqslant n$. Then the condition of the problem can be written as

$$
\frac{S_{n}-m}{n-1}=40 \frac{3}{4}
$$

Transforming: $\frac{n+2}{2}-\frac{m-1}{n-1}=40 \frac{3}{4}$. Note that $\frac{m-1}{n-1} \leqslant 1$, so if $n$ is even, then $\frac{m-1}{n-1}=\frac{1}{4}$, and if odd, then $\frac{m-1}{n-1}=\frac{3}{4}$. In the first case, we get $n=80$, from which $m-1=79 \cdot \frac{1}{4}-$ not an integer. In the second case, $n=81$, i.e., $m-1=80 \cdot \frac{3}{4}=60$.