1. [5 points] $S$ is the sum of the first 15 terms of an increasing arithmetic progression $a_{1}, a_{2}, a_{3}, \ldots$, consisting of integers. It is known that $a_{7} a_{16}>S-24, a_{11} a_{12}<S+4$. Determine all possible values of $a_{1}$.
Answer: $-5; -4; -2; -1$.

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

$$
\left\{\begin{array} { l } 
{ ( a _ { 1 } + 6 d ) ( a _ { 1 } + 1 5 d ) > S - 2 4 , } \\
{ ( a _ { 1 } + 1 0 d ) ( a _ { 1 } + 1 1 d ) S-24 \\
a_{1}^{2}+21 a_{1} d+110 d^{2} 1 5 a _ { 1 } + 1 0 5 - 2 4 , } \\
{ a _ { 1 } ^ { 2 } + 2 1 a _ { 1 } + 1 1 0  0 , } \\
{ a _ { 1 } ^ { 2 } + 6 a _ { 1 } + 1 < 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
a_{1} \neq-3 \\
a_{1} \in(-3-\sqrt{8} ;-3+\sqrt{8})
\end{array}\right.\right.\right.
$$

Since $a_{1} \in \mathbb{Z}$, then $a_{1} \in\{-5; -4; -2; -1\}$.