ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 531

\[\boxed{\mathbf{531}.}\]

\[S = 3,5;\ \ S_{2} = \frac{147}{16}:\]

\[\frac{S_{2}}{S} = \frac{b_{1}^{2}}{1 - q^{2}}\ :\frac{b_{1}}{1 - q} = \frac{b_{1}}{1 + q}.\]

\[\left\{ \begin{matrix} \frac{b_{1}}{1 - q} = \frac{7}{2}\text{\ \ \ \ \ \ \ \ \ \ \ } \\ \frac{b_{1}}{1 + q} = \frac{147}{16}\ :\frac{7}{2} \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} b_{1} = \frac{7}{2}(1 - q)\text{\ \ \ } \\ b_{1} = \frac{21}{8}(1 + q) \\ \end{matrix} \right.\ \]

\[\frac{7}{2}(1 - q) = \frac{21}{8}(1 + q)\]

\[1 - q = \frac{3}{4}(1 + q)\]

\[1 - q = \frac{3}{4} + \frac{3}{4}q\]

\[q = \frac{1}{7}.\]

\[b_{1} = \frac{21}{8}\left( 1 + \frac{1}{7} \right) = \frac{21}{8} \cdot \frac{8}{7} = 3;\]

\[S_{3} = \frac{b_{1}^{3}}{1 - q^{3}} = 3^{3}\ :\left( 1 - \frac{1}{343} \right) =\]

\[= \frac{27 \cdot 343}{342} = \frac{1029}{38}.\]

\[Ответ:\frac{1029}{38}.\]