Решебник по алгебре 9 класс Макарычев Задание 578

\[\boxed{\text{578}\text{\ (578)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ a_{1} = - 3;\ \ d = 0,7:\]

\[a_{n} = - 3 + 0,7 \cdot (n - 1)\]

\[a_{11} = - 3 + 0,7 \cdot 10 = 4.\]

\[\textbf{б)}\ a_{1} = 18;\ \ d = - 0,6:\ \]

\[a_{n} = 18 - 0,6 \cdot (n - 1)\]

\[a_{26} = 18 - 0,6 \cdot 25 = 3.\]

\[\boxed{\text{578.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[c_{7} = c_{1} + 6d = 18,5;\ \ \ \]

\[\ c_{17} = c_{1} + 16d = - 26,5:\]

\[\left\{ \begin{matrix} c_{1} + 6d = 18,5\ \ \ \ \ \ \\ c_{1} + 16d = - 26,5 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} c_{1} = 18,5 - 6d\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 18,5 - 6d + 16d = - 26,5 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} 10d = - 45\ \ \ \ \ \ \ \\ c_{1} = 18,5 - 6d \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} d = - 4,5 \\ c_{1} = 45,5 \\ \end{matrix} \right.\ \]

\[S_{20} = \frac{2c_{1} + d(n - 1)}{2} \cdot n =\]

\[= \frac{2 \cdot 45,5 - 4,5 \cdot (20 - 1)}{2} \cdot 20 =\]

\[= (91 - 85,5) \cdot 10 = 55.\]