1. Find the sum of the first thirty-three terms of an arithmetic progression, given that the sum of the seventh, eleventh, twentieth, and thirtieth terms of this progression is 28.
Answer. $S_{33}=231$.

Solution Let $a_{k}$ be the $k$-th term of the arithmetic progression, and $d$ be its common difference. Then, according to the condition, $a_{7}+a_{11}+a_{20}+a_{30}=28$, from which we have $a_{7}+\left(a_{7}+4 d\right)+\left(a_{7}+13 d\right)+\left(a_{7}+23 d\right)=28 \Leftrightarrow a_{7}+10 d=7$, i.e., $a_{17}=7$. The sum of the first thirty-three terms is

$$
S_{33}=\frac{a_{1}+a_{33}}{2} \cdot 33=\frac{\left(a_{17}-16 d\right)+\left(a_{17}+16 d\right)}{2} \cdot 33=33 a_{17}=231
$$