Найдите первый член и разность арифметической прогрессии (an), если: a5+a9=42 и a3*a10=165.

\[a_{5} + a_{9} = 42;\ \ \ \ \ \ a_{3} \cdot a_{10} = 165\]

\[\left\{ \begin{matrix} a_{5} + a_{9} = 42\ \ \ \\ a_{3} \cdot a_{10} = 165 \\ \end{matrix}\text{\ \ \ \ \ \ \ \ \ \ } \right.\ \]

\[\left\{ \begin{matrix} a_{1} + 4d + a_{1} + 8d = 42\ \ \ \ \\ \left( a_{1} + 2d \right)\left( a_{1} + 9d \right) = 165 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} 2a_{1} + 12d = 42\ \ \ \ \ \ \ \ \ \ \ \ \ |\ :2 \\ \left( a_{1} + 2d \right)\left( a_{1} + 9d \right) = 165\ \\ \end{matrix}\text{\ \ \ \ \ \ \ \ \ } \right.\ \]

\[(21 - 4d)(21 + 3d) = 165\]

\[4d^{2} + 7d - 92 = 0\]

\[D = 49 + 1472 = 1521\]

\[d_{1} = \frac{- 7 + 39}{8} = 4,\ \ \ \ \ \ \ \ \ \ \]

\[d_{2} = \frac{- 7 - 39}{8} = - 5,75\]

\[\left\{ \begin{matrix} d_{1} = 4\ \ \ \ \ \ \ \ \ \ \ \ \\ a_{1} = 21 - 24 \\ \end{matrix}\text{\ \ \ \ \ \ \ \ } \right.\ \text{\ \ }\]

\[\left\{ \begin{matrix} d = - 5,75\ \ \ \ \ \ \ \ \\ a_{1} = 21 + 34,5 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} d_{1} = 4\ \ \ \\ a_{1} = - 3 \\ \end{matrix}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \right.\ \text{\ \ }\]

\[\left\{ \begin{matrix} d = - 5,75 \\ a_{1} = 55,5\ \\ \end{matrix} \right.\ \]

\[Ответ:a_{1} = - 3,\ \ \ d = 4\ ;\ \ \ \]

\[a_{1} = 55,5,\ \ \ d = - 5,75.\]