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 is to the sum of all terms except the last three as $4: 3$. Find the number of terms in this progression.
Answer: 20.

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:

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

or, after transformations,

$$
\left\{\begin{aligned}
a & =(n-19) d \\
-2 a & =(n-22) d
\end{aligned}\right.
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_ec0b1918074dfb35bb6bg-02.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 $n=20$.