Problem 1. The sum of the first thirteen terms of a certain arithmetic progression is $50 \%$ of the sum of the last thirteen terms of this progression. The sum of all terms of this progression except the first three terms is in the ratio $6: 5$ to the sum of all terms except the last three. Find the number of terms in this progression.
Answer: 24.

Solution. Let $a$ be the first term of the arithmetic progression, $d$ be its common difference, and $n$ be the number of terms. Then, the sum of the first thirteen terms is $\frac{13 \cdot(a+(a+12 d))}{2}$, the sum of the last thirteen terms is $-\frac{13 \cdot(a+(n-13) d+a+(n-1) d)}{2}$, the sum of all terms except the first three is $-\frac{(n-3) \cdot(a+3 d+a+(n-1) d)}{2}$, and the sum of all terms except the last three is $-\frac{(n-3) \cdot(a+a+(n-4) d)}{2}$. From the condition, we have the system:

$$
\left\{\begin{aligned}
2 \cdot 13(a+6 d) & =13(a+(n-7) d) \\
5 \cdot(n-3) \cdot(2 a+(n+2) d) & =6 \cdot(n-3) \cdot(2 a+(n-4) d)
\end{aligned}\right.
$$

or, after transformations,

$$
\left\{\begin{aligned}
a & =(n-19) d \\
-2 a & =(n-34) d
\end{aligned}\right.
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_ec0b1918074dfb35bb6bg-11.jpg?height=52&width=1879&top_left_y=1179&top_left_x=91)
(since otherwise the sum of all terms except the first three would equal the sum of all terms except the last three), we get $n=24$.