Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 183

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\[1)\ q = \frac{1}{2};\ \ S_{6} = - 15\frac{3}{4} = - \frac{63}{4}:\]

\[S_{6} = \frac{b_{1} \cdot \left( 1 - q^{6} \right)}{1 - q} =\]

\[= \frac{b_{1} \cdot \left( 1 - \left( \frac{1}{2} \right)^{6} \right)}{1 - \frac{1}{2}} =\]

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

\[= 2b_{1} \cdot \frac{63}{64} = \frac{63b_{1}}{32}\]

\[\frac{63b_{1}}{32} = - \frac{63}{4}\]

\[b_{1} = - \frac{63}{4} \cdot \frac{32}{63} = - 8.\]

\[Ответ:b_{1} = - 8.\]

\[2)\ q = - \frac{1}{2};\ \ S_{5} = 4\frac{1}{8} = \frac{33}{8}:\]

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

\[= \frac{b_{1} \cdot \left( 1 - \left( - \frac{1}{2} \right)^{5} \right)}{1 + \frac{1}{2}} =\]

\[= \frac{b_{1} \cdot \left( 1 + \frac{1}{32} \right)}{\frac{3}{2}} = \frac{2b_{1} \cdot \frac{33}{32}}{3} =\]

\[= \frac{b_{1} \cdot 33}{3 \cdot 16} = \frac{11b_{1}}{16\ }\]

\[\frac{11b_{1}}{16\ } = \frac{33}{8}\]

\[b_{1} = \frac{33}{8} \cdot \frac{16}{11} = 3 \cdot 2 = 6.\]

\[Ответ:b_{1} = 6.\]