Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 432

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\[\frac{S_{3}}{S_{2}} = 3;\ \ \frac{S}{S_{2}} = \frac{3}{7}:\]

\[\frac{S_{3}}{S_{2}} = \frac{b_{1}^{3}}{1 - q^{3}}\ \ :\frac{b_{1}^{2}}{1 - q^{2}} =\]

\[= \frac{b_{1}^{3}(1 - q)(1 + q)}{(1 - q)\left( 1 + q + q^{2} \right) \cdot b_{1}^{2}} =\]

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

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

\[= \frac{b_{1}(1 - q)(1 + q)}{(1 - q)b_{1}^{2}} = \frac{3}{7};\]

\[b_{1} = \frac{7}{3}(1 + q).\]

\[{\frac{7}{3}(1 + q)(1 + q) = }{= 3 \cdot \left( q^{2} + q + 1 \right)\ \ | \cdot 3}\]

\[7 \cdot \left( 1 + 2q + q^{2} \right) =\]

\[= 9 \cdot \left( q^{2} + q + 1 \right)\]

\[2q^{2} - 5q + 2 = 0\]

\[D = 25 - 16 = 9\]

\[q_{1} = \frac{5 + 3}{4} = 2;\ \ \ \]

\[q_{2} = - \frac{5 - 3}{4} = \frac{1}{2}\]

\[Прогрессия\ бесконечно\ \]

\[убывающая:\ \ q = \frac{1}{2}.\]

\[b_{1} = \frac{7}{3}\left( 1 + \frac{1}{2} \right) = \frac{7}{3} + \frac{7}{6} =\]

\[= \frac{14}{6} + \frac{7}{6} = \frac{21}{6} = 3,5.\]