1. [5 points] $S$ - the sum of the first 9 terms of an increasing arithmetic progression $a_{1}, a_{2}, a_{3}, \ldots$, consisting of integers. It is known that $a_{5} a_{18}>S-4, a_{13} a_{10}<S+60$. Indicate all possible values of $a_{1}$.
Answer: $-10; -9; -8; -7; -5; -4; -3; -2$.

Solution. Let the common difference of the progression be $d$. The inequalities given in the condition can be transformed as follows:

$$
\left\{\begin{array} { l } 
{ ( a _ { 1 } + 4 d ) ( a _ { 1 } + 1 7 d ) > S - 4 , } \\
{ ( a _ { 1 } + 1 2 d ) ( a _ { 1 } + 9 d ) S-4 \\
a_{1}^{2}+21 a_{1} d+108 d^{2} 9 a _ { 1 } + 3 6 - 4 , } \\
{ a _ { 1 } ^ { 2 } + 2 1 a _ { 1 } + 1 0 8  0 , } \\
{ a _ { 1 } ^ { 2 } + 1 2 a _ { 1 } + 1 2 < 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
a_{1} \neq-6 \\
a_{1} \in(-6-\sqrt{24} ;-6+\sqrt{24})
\end{array}\right.\right.\right.
$$

Since $a_{1} \in \mathbb{Z}$, then $a_{1} \in\{-10; -9; -8; -7; -5; -4; -3; -2\}$.