1. [5 points] $S$ - the sum of the first 6 terms of an increasing arithmetic progression $a_{1}, a_{2}, a_{3}, \ldots$, consisting of integers. It is known that $a_{10} a_{16}>S+39, a_{11} a_{15}<S+55$. Indicate all possible values of $a_{1}$.
Answer: $-12; -11; -10; -8; -7; -6$.

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

$$
\left\{\begin{array} { l } 
{ ( a _ { 1 } + 9 d ) ( a _ { 1 } + 1 5 d ) > S + 3 9 , } \\
{ ( a _ { 1 } + 1 0 d ) ( a _ { 1 } + 1 4 d ) S+39 \\
a_{1}^{2}+24 a_{1} d+140 d^{2} 6 a _ { 1 } + 1 5 + 3 9 , } \\
{ a _ { 1 } ^ { 2 } + 2 4 a _ { 1 } + 1 4 0  0 , } \\
{ a _ { 1 } ^ { 2 } + 1 8 a _ { 1 } + 7 0 < 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
a_{1} \neq-9 \\
a_{1} \in(-9-\sqrt{11} ;-9+\sqrt{11})
\end{array}\right.\right.\right.
$$

Since $a_{1} \in \mathbb{Z}$, then $a_{1} \in\{-12; -11; -10; -8; -7; -6\}$.