Решебник по алгебре 9 класс Макарычев Задание 585

\[\boxed{\text{585\ (}\text{с}\text{).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[y_{n} = y_{1} + d(n - 1)\text{\ \ }\]

\[y_{n} - y_{1} = d(n - 1)\]

\[d = \frac{y_{n} - y_{1}}{n - 1}\]

\[\textbf{а)}\ y_{1} = 10;\ \ y_{5} = 22:\ \]

\[d = \frac{22 - 10}{4} = \frac{12}{4} = 3.\]

\[\textbf{б)}\ y_{1} = 28;\ \ y_{15} = - 21:\]

\[d = \frac{- 21 - 28}{14} = - 3,5.\]

\[\boxed{\text{585.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[a_{7} = a_{1} + 6d = 8,\ \ \]

\[a_{11} = a_{1} + 10d = 12,8;\]

\[\Longrightarrow \left\{ \begin{matrix} a_{1} + 6d = 8\ \ \ \ \ \ \\ a_{1} + 10d = 12,8 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} a_{1} = 8 - 6d\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 8 - 6d + 10d = 12,8 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} 4d = 4,8\ \ \ \ \ \\ a_{1} = 8 - 6d \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} d = 1,2\ \ \\ a_{1} = 0,8. \\ \end{matrix} \right.\ \]

\[Ответ:a_{1} = 0,8;\ \ \ d = 1,2.\]