1. Find the sum of all irreducible fractions with denominator 7 that lie between the natural numbers $a$ and $b, a<b$.
Solution. Let's write $a$ and $b$ in the form $\frac{7 a}{7}$ and $\frac{7 b}{7}$. All fractions with a denominator equal to 7, including $a$ and $b$, can be written as

$$
\frac{7 a}{7}, \frac{7 a+1}{7}, \frac{7 a+2}{7}, \ldots ., \frac{7 b-1}{7}, \frac{7 b}{7}
$$

Of these, only

$$
\frac{7 a}{7}, \frac{7 a+7}{7}, \frac{7 a+14}{7}, \ldots ., \frac{7 b-7}{7}, \frac{7 b}{7}
$$

are reducible. The sum of the irreducible fractions will be calculated by subtracting the sum of the numbers in sequence (2) from the sum of all fractions in sequence (1).

The sum of the numbers in sequence (1) is the sum of numbers from an arithmetic progression with the first term $\frac{7 a}{7}$ and difference $\frac{1}{7}$. The total number of terms will be determined from the equation

$$
\begin{aligned}
& b=a+(n-1) \frac{1}{7} \\
& 7 b=7 a+n-1 \\
& n=7 b-7 a+1
\end{aligned}
$$

Thus, the sum of the numbers in the arithmetic progression (1) is:

$$
\frac{n}{2}\left(2 a_{1}+(n-1) d\right)=\frac{7 b-7 a+1}{2}\left[2 a+(7 b-7 a+1-1) \frac{1}{7}\right]=\frac{7 b-7 a+1}{2}(a+b) .
$$

The numbers in sequence (2) are also terms of an arithmetic progression where the difference is equal to 1 and the last term is $b$, so the number of its terms is

$$
\begin{aligned}
& b=a+(k-1) \cdot 1 \\
& k=b-a+1
\end{aligned}
$$

Thus, their sum is equal to

$$
\frac{b-a+1}{2}(2 a+(b-a+1-1))=\frac{b-a+1}{2}(a+b) .
$$

Finally, the desired sum is:

$$
\begin{aligned}
S & =\frac{(b+a)(7 b-7 a+1)}{2}-\frac{(b+a)(b-a+1)}{2}=\frac{b+a}{2}(7 b-7 a+1-b+a-1) \\
& =\frac{b+a}{2}(6 a-6 b)=3\left(a^{2}-b^{2}\right)
\end{aligned}
$$