Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 530

\[\boxed{\mathbf{530}.}\]

\[1)\ S = \frac{2}{3};\ \ S_{5} = \frac{31}{48}:\]

\[S_{5} = \frac{b_{1}\left( 1 - q^{5} \right)}{1 - q} = \frac{31}{48};\]

\[S = \frac{b_{1}}{1 - q} = \frac{2}{3};\]

\[b_{1} = \frac{2}{3}(1 - q);\]

\[S_{5} = \frac{2 \cdot \left( 1 - q^{5} \right)}{3} = \frac{31}{48}\]

\[96 \cdot \left( 1 - q^{5} \right) = 93\]

\[1 - q^{5} = \frac{93}{96}\]

\[q^{5} = 1 - \frac{93}{96} = \frac{1}{32}\]

\[q^{5} = \left( \frac{1}{2} \right)^{5} \Longrightarrow q = \frac{1}{2};\]

\[b_{1} = \frac{2}{3}\left( 1 - \frac{1}{2} \right) = \frac{1}{3}.\]

\[Ответ:b_{1} = \frac{1}{2};\ \ q = \frac{1}{2}\text{.\ }\]

\[2)\ S = 0,9;\ \ S_{8} = \frac{8}{9}:\]

\[S = \frac{b_{1}}{1 - q};\ \ \]

\[\ S_{8} = \frac{b_{1}\left( 1 - q^{8} \right)}{1 - q} = \frac{8}{9};\]

\[\left( 1 - q^{8} \right) \cdot 0,9 = \frac{8}{9}\ \ \ | \cdot \frac{10}{9}\]

\[1 - q^{8} = \frac{80}{81}\]

\[q^{8} = \frac{1}{81}\]

\[q^{2} = \frac{1}{3}\]

\[q = \pm \frac{\sqrt{3}}{3}.\]

\[b_{1} = 0,9 \cdot (1 - q) =\]

\[= 0,9 \cdot \left( 1 \pm \frac{\sqrt{3}}{3} \right) = \frac{9 \cdot (\sqrt{3} \pm 1)}{10\sqrt{3}}.\]