Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 431

\[\boxed{\mathbf{431}.}\]

\[b_{2} = 6;\ \]

\[\ S = \frac{1}{8} \cdot \underset{S_{кв}}{\overset{\left( b_{1}^{2} + b_{2}^{2} + \ldots + b_{n}^{2} \right)}{︸}}\]

\[S_{кв} = \frac{b_{1}^{2}}{1 - q²} \rightarrow \frac{S_{кв}}{S} = 8.\]

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

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

\[= \frac{b_{1}}{1 + q} = 8\]

\[b_{1} = 8 \cdot (1 + q);\ \ \ \]

\[b_{2} = b_{1} \cdot q \rightarrow b_{1} = \frac{b_{2}}{q} = \frac{6}{q}.\]

\[8 \cdot (1 + q) = \frac{6}{q}\]

\[4q(1 + q) = 3\]

\[4q^{2} + 4q - 3 = 0\]

\[D_{1} = 4 + 12 = 16\]

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

\[\text{\ \ }q_{2} = \frac{- 2 - 4}{4} =\]

\[= - 1,5\ (не\ подходит).\]

\[b_{1} = 6\ :\frac{1}{2} = 12.\]

\[Ответ:b_{1} = 12;\ \ q = \frac{1}{2}.\]