Problem 1. The sum of the first three terms of an arithmetic progression, as well as the sum of the first six terms, are natural numbers. In addition, its first term $d_{1}$ satisfies the inequality $d_{1} \geqslant \frac{1}{2}$. What is the smallest value that $d_{1}$ can take?
Answer: $5 / 9$

Solution. Let $d_{n}$ be the $n$-th term of the progression, $d$ be the common difference of the progression, and $S_{n}=d_{1}+d_{2}+\ldots+d_{n}$ be the sum of the first $n$ terms of the progression.

Express $d_{1}$ in terms of $S_{3}$ and $S_{6}$. Note that $S_{3}=3 d_{1}+3 d, S_{6}=6 d_{1}+15 d$, from which $d_{1}=\frac{5 S_{3}-S_{6}}{9}$. By the condition $\frac{5 S_{3}-S_{6}}{9} \geqslant \frac{1}{2}$. Therefore, $5 S_{3}-S_{6} \geqslant \frac{9}{2}$. Since $S_{3}$ and $S_{6}$ are natural numbers by the condition, the smallest value of $5 S_{3}-S_{6}$ is 5 (it is achieved, for example, when $S_{3}=2, S_{6}=5$). Therefore, $\min d_{1}=\frac{5}{9}$.