2. Several numbers form an arithmetic progression, where their sum is 63, and the first term is one and a half times the common difference of the progression. If all terms of the progression are decreased by the same amount so that the first term of the progression is equal to the common difference of the progression, then the sum of all numbers will decrease by no more than 8, but no less than 7. Determine what the common difference of this progression can be.
# Problem 2.

Answer: the difference is $21 / 8$ or 2.

Solution. Let the numbers $a_{1}, a_{2}, \ldots, a_{n}$ form an arithmetic progression with a common difference $d$. Then $a_{1}+a_{2}+\cdots+a_{n}=63, a_{1}=3 d / 2 \Rightarrow 0.5 d(n+2) n=63$.

After the described change in the condition, these numbers will again form an arithmetic progression with the same difference $d$, but with a different first term, now equal to $d$. Therefore, their sum will be equal to $0.5 d(n+1) n$, and according to the condition, the inequality $55 \leqslant 0.5 d(n+1) n \leqslant 56$ is satisfied. Considering the first relation, we get

$$
\frac{55}{63} \leqslant \frac{n+1}{n+2} \leqslant \frac{56}{63} \Longleftrightarrow \frac{47}{8} \leqslant n \leqslant 7
$$

Therefore, $n=6$ or $n=7$, and the common difference $d$ is respectively equal to $21 / 8$ or 2.