1. [5 points] $S$ - the sum of the first 5 terms of an increasing arithmetic progression $a_{1}, a_{2}, a_{3}, \ldots$, consisting of integers. It is known that $a_{6} a_{11}>S+15, a_{9} a_{8}<S+39$. Indicate all possible values of $a_{1}$.
Answer: $-9; -8; -7; -6; -4; -3; -2; -1$.

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

$$
\left\{\begin{array} { l } 
{ ( a _ { 1 } + 5 d ) ( a _ { 1 } + 10 d ) > S + 15 , } \\
{ ( a _ { 1 } + 8 d ) ( a _ { 1 } + 7 d ) S+15 \\
a_{1}^{2}+15 a_{1} d+56 d^{2} > 5 a _ { 1 } + 10 + 15 , } \\
{ a _ { 1 } ^ { 2 } + 15 a _ { 1 } + 56 > 0 , } \\
{ a _ { 1 } ^ { 2 } + 10 a _ { 1 } + 7 < 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
a_{1} \neq -5 \\
a_{1} \in (-5-\sqrt{18} ; -5+\sqrt{18})
\end{array}\right.\right.\right.
$$

Since $a_{1} \in \mathbb{Z}$, then $a_{1} \in \{-9; -8; -7; -6; -4; -3; -2; -1\}$.