Сумма восьмого и шестого членов арифметической прогрессии равна 16, а произведение второго и двенадцатого равно -36. Найдите разность и первый член данной прогрессии.

\[\left\{ \begin{matrix} a_{8} + a_{6} = 16\ \ \ \\ a_{2} \cdot a_{12} = - 36 \\ \end{matrix} \right.\ \]

\[a_{2} = a_{1} + d\ \ \ \ \ \]

\[a_{6} = a_{1} + 5d\]

\[a_{8} = a_{1} + 7d\ \ \ \ \ \ \]

\[a_{12} = a_{1} + 11d\]

\[\left\{ \begin{matrix} a_{1} + 7d + a_{1} + 5d = 16\ \ \ \ \\ \left( a_{1} + d \right)\left( a_{1} + 11d \right) = - 36 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} 2a_{1} + 12d = 16\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ a_{1}^{2} + 11a_{1}d + a_{1}d + 11d^{2} = - 36 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} a_{1} + 6d = 8 \longrightarrow a_{1} = 8 - 6d\ \\ a_{1}^{2} + 12a_{1}d + 11d^{2} = - 36\ \ \ \ \\ \end{matrix} \right.\ \]

\[- 25d^{2} = - 100\ \ \ \ \ \]

\[d^{2} = 4\ \ \ \ \ \]

\[d = \pm 2.\]

\[a_{1} = 8 - 6 \cdot ( \pm 2) = 8 \pm 12 =\]

\[= 20;\ \ - 4\]

\[Ответ:\ \left\{ \begin{matrix} a_{1} = 20 \\ d = - 2 \\ \end{matrix} \right.\ \text{\ \ \ }или\ \]

\[\text{\ \ }\left\{ \begin{matrix} a_{1} = - 4 \\ d = 2\ \ \ \ \ \\ \end{matrix} \right.\ .\]