Решебник по алгебре 9 класс Макарычев Задание 696

\[\boxed{\text{696}\text{\ (696)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[S_{10} = 100,\ \ S_{30} = 900,\]

\[S_{10} = \frac{2a_{1} + 9d}{2} \cdot 10 = 5 \cdot \left( 2a_{1} + 9d \right) = 100 \Longrightarrow 2a_{1} + 9d = 20,\]

\[S_{30} = \frac{2a_{1} + 29d}{2} \cdot 30 = 15 \cdot \left( 2a_{1} + 29d \right) = 900 \Longrightarrow 2a_{1} + 29d = 60,\]

\[\left\{ \begin{matrix} 2a_{1} + 9d = 20\ \ \\ 2a_{1} + 29d = 60 \\ \end{matrix} \right.\ -\]

\[\left\{ \begin{matrix} 2a_{1} + 9d = 20 \\ 20d = 40\ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} d = 2 \\ a_{1} = 1 \\ \end{matrix} \right.\ \]

\[S_{40} = \frac{2a_{1} + 39d}{2} \cdot 40 = 20 \cdot (2 + 39 \cdot 2) = 1600.\]

\[\boxed{\text{696.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[x_{2} = - 32,\ \ q = - \frac{1}{2},\]

\[\text{\ \ }S_{10} = ?\]

\[x_{2} = x_{1} \cdot q,\ \]

\[x_{1} = \frac{x_{2}}{q} = 32 \cdot 2 = 64,\]

\[S_{10} = \frac{x_{1} \cdot \left( q^{10} - 1 \right)}{q - 1} =\]

\[= \frac{64 \cdot \left( \left( - \frac{1}{2} \right)^{10} - 1 \right)}{- \frac{1}{2} - 1\ } =\]

\[= \frac{64 \cdot 1023}{1024} \cdot \frac{2}{3} = \frac{341}{8} = 42\frac{5}{8}.\]