1. [5 points] $S$ is the sum of the first 10 terms of an increasing arithmetic progression $a_{1}, a_{2}, a_{3}, \ldots$, consisting of integers. It is known that $a_{6} a_{12}>S+1, a_{7} a_{11}<S+17$. Determine all possible values of $a_{1}$.
Answer: $-6; -5; -4; -2; -1; 0$.

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

$$
\left\{\begin{array} { l } 
{ ( a _ { 1 } + 5 d ) ( a _ { 1 } + 1 1 d ) > S + 1 , } \\
{ ( a _ { 1 } + 6 d ) ( a _ { 1 } + 1 0 d ) S+1 \\
a_{1}^{2}+16 a_{1} d+60 d^{2} 1 0 a _ { 1 } + 4 5 + 1 , } \\
{ a _ { 1 } ^ { 2 } + 1 6 a _ { 1 } + 6 0  0 , } \\
{ a _ { 1 } ^ { 2 } + 6 a _ { 1 } - 2 < 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
a_{1} \neq-3 \\
a_{1} \in(-3-\sqrt{11} ;-3+\sqrt{11})
\end{array}\right.\right.\right.
$$

Since $a_{1} \in \mathbb{Z}$, then $a_{1} \in\{-6; -5; -4; -2; -1; 0\}$.