Решебник по алгебре 9 класс Мерзляк Задание 800

\[\boxed{\mathbf{800\ (800).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[S_{6} = 39;\ \ S_{14} = - 77\]

\[S_{6} = \frac{2a_{1} + 5d}{2} \cdot 6 =\]

\[= \left( 2a_{1} + 5d \right) \cdot 3\]

\[S_{14} = \frac{2a_{1} + 13d}{2} \cdot 14 =\]

\[= \left( 2a_{1} + 13d \right) \cdot 7\]

\[\left\{ \begin{matrix} \left( 2a_{1} + 5d \right) \cdot 3 = 39\ \ \ \ \ \ \ \ \ |\ :3 \\ \left( 2a_{1} + 13d \right) \cdot 7 = - 77\ \ \ |\ :7 \\ \end{matrix} \right.\ \text{\ \ \ \ }\]

\[\text{\ \ \ \ \ \ }\left\{ \begin{matrix} 2a_{1} + 5d = 13\ \ \ \ \ \ \\ 2a_{1} + 13d = - 11 \\ \end{matrix} \right.\ \ ( - )\]

\[\text{\ \ }\left\{ \begin{matrix} - 8d = 24\ \ \ \ \ \ \ \\ a_{1} = \frac{13 - 5d}{2} \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} d = - 3\ \ \ \ \ \ \ \ \ \ \ \\ a_{1} = \frac{13 + 15}{2} \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\left\{ \begin{matrix} d = - 3 \\ a_{1} = 14 \\ \end{matrix} \right.\ \]

\[Ответ:\ a_{1} = 14;\ \ d = - 3.\]