Решебник по алгебре 9 класс Макарычев Задание 590

\[\boxed{\text{590}\text{\ (590)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

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

\[x_{16} = - 7;\ \ x_{26} = 55:\]

\[\left\{ \begin{matrix} x_{1} + 15d = - 7 \\ x_{1} + 25d = 55 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} 10d = 62\ \ \ \ \ \ \ \ \ \\ x_{1} = - 7 - 15d \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} d = 6,2\ \ \ \ \ \ \ \\ x_{1} = - 100. \\ \end{matrix} \right.\ \]

\[Ответ:d = 6,2;\ \ x_{1} = - 100.\]

\[\boxed{\text{590.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[c_{n} = c_{1} \cdot q^{n - 1};\]

\[\textbf{а)}\ c_{6} = c_{1} \cdot q^{5};\]

\[\textbf{б)}\ c_{20} = c_{1} \cdot q^{19};\]

\[\textbf{в)}\ c_{125} = c_{1} \cdot q^{124};\ \]

\[\textbf{г)}\ c_{k} = c_{1} \cdot q^{k - 1};\]

\[\textbf{д)}\ c_{k + 3} = c_{1} \cdot q^{k + 2};\]

\[\textbf{е)}\ c_{2k} = c_{1} \cdot q^{2k - 1}.\]