Решебник по алгебре 9 класс Мерзляк Задание 771

\[\boxed{\mathbf{771\ (771).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ a_{1} = 6;\ \ a_{9} = 22\]

\[S_{12} = \frac{2a_{1} + 11d}{2} \cdot 12 =\]

\[= \left( 2a_{1} + 11d \right) \cdot 6\]

\[a_{9} = a_{1} + 8d = 22\ \ \]

\[8d = 22 - 6\]

\[8d = 16\ \]

\[d = 2\]

\[S_{12} = (2 \cdot 6 + 11 \cdot 2) \cdot 6 =\]

\[= (12 + 22) \cdot 6 = 34 \cdot 6 = 204.\]

\[Ответ:204.\]

\[2)\ a_{6} = 49;\ \ a_{20} = 7\]

\[\left\{ \begin{matrix} a_{6} = a_{1} + 5d = 49\ \\ a_{20} = a_{1} + 19d = 7 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} a_{1} + 5d = 49 \\ a_{1} + 19d = 7 \\ \end{matrix} \right.\ ( - )\text{\ \ \ }\]

\[\text{\ \ }\left\{ \begin{matrix} - 14d = 42\ \ \ \ \\ a_{1} = 49 - 5d \\ \end{matrix} \right.\ \text{\ \ \ \ }\]

\[\left\{ \begin{matrix} d = - 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ a_{1} = 49 + 5 \cdot 3 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} d = - 3\ \ \ \ \ \ \ \ \ \ \ \\ a_{1} = 49 + 15 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} d = - 3\ \\ a_{1} = 64 \\ \end{matrix} \right.\ \]

\[S_{12} = \left( 2 \cdot 64 + 11 \cdot ( - 3) \right) \cdot 6 =\]

\[= (128 - 33) \cdot 6 = 570.\]

\[Ответ:570.\]