Найдите первый член и разность арифметической прогрессии (an), если: a4+a10=16 и a2*a6=-12.

Ответ:

\[a_{4} + a_{10} = 16;\ \ a_{2} + a_{6} = - 12:\ \]

\[\left\{ \begin{matrix} a_{1} + 3d + a_{1} + 9d = 16 \\ \left( a_{1} + d \right)\left( a_{1} + 5d \right) = - 12 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} 2a_{1} + 12d = 16\ \ |\ :2\ \ \ \ \ \ \ \ \\ \left( a_{1} + d \right)\left( a_{1} + 5d \right) = - 12 \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} a_{1} = 8 - 6d\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ (8 - 5d)(8 - d) = - 12 \\ \end{matrix} \right.\ \]

\[64 - 8d + 40d + 5d^{2} + 12 = 0\]

\[5d^{2} - 48d + 76 = 0\]

\[D = 2304 - 1520 = 784\]

\[d_{1} = \frac{48 - 28}{10} = 2;\ \ \]

\[d_{2} = \frac{48 + 28}{10} = 7,6\]

\[a_{1} = - 4;\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a_{2} = - 37,6\]

\[Ответ:\ \ d = 2;\ \ a = - 4\ \ \ \]

\[или\ \ \ d = 7,6;\ \ a = - 37,6.\]