Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 528

\[\boxed{\mathbf{528}.}\]

\[\left( b_{n} \right) - данная\ геометрическая\ \]

\[прогрессия\ со\ знаменателем\ q;\]

\[\mathbf{Сумма\ первых\ трех\ членов\ }\]

\[равна\ 39:\]

\[b_{1} + b_{2} + b_{3} = 39;\]

\[b_{1} + b_{1} \bullet q + b_{1} \bullet q^{2} = 39;\]

\[b_{1} \bullet \left( 1 + q + q^{2} \right) = 39;\]

\[b_{1} = \frac{39}{1 + q + q^{2}}.\]

\[Сумма\ обратных\ им\ величин\ \]

\[равна\ \frac{13}{27}:\]

\[\frac{1}{b_{1}} + \frac{1}{b_{2}} + \frac{1}{b_{3}} = \frac{13}{27};\]

\[\frac{1}{b_{1}} + \frac{1}{b_{1} \bullet q} + \frac{1}{b_{1} \bullet q^{2}} = \frac{13}{27};\]

\[\frac{q^{2} + q + 1}{b_{1} \bullet q^{2}} = \frac{13}{27};\]

\[\frac{\left( q^{2} + q + 1 \right)^{2}}{39 \bullet q^{2}} = \frac{13}{27};\]

\[\left( \frac{q^{2} + q + 1}{q} \right)^{2} = \frac{13 \bullet 39}{27} = \frac{169}{9};\]

\[\frac{q^{2} + q + 1}{q} = \sqrt{\frac{169}{3}} = \frac{13}{3};\]

\[3\left( q^{2} + q + 1 \right) = 13q;\]

\[3q^{2} + 3q + 3 - 13q = 0;\]

\[3q^{2} - 10q + 3 = 0;\]

\[D = 10^{2} - 4 \bullet 3 \bullet 3 =\]

\[= 100 - 36 = 64\]

\[q_{1} = \frac{10 - 8}{2 \bullet 3} = \frac{2}{6} = \frac{1}{3};\]

\[q_{2} = \frac{10 + 8}{2 \bullet 3} = \frac{18}{6} = 3.\]

\[Дана\ бесконечно\ убывающая\]

\[\ прогрессия:\]

\[q = \frac{1}{3};\]

\[b_{1} = \frac{39}{1 + \frac{1}{3} + \frac{1}{9}} =\]

\[= 39\ :\left( \frac{9}{9} + \frac{3}{9} + \frac{1}{9} \right) = 39\ :\frac{13}{9} =\]

\[= 39 \bullet \frac{9}{13} = 3 \bullet 9 = 27;\]

\[S = \frac{b_{1}}{1 - q} = \frac{27}{1 - \frac{1}{3}} = 27\ :\frac{2}{3} =\]

\[= 27 \bullet \frac{3}{2} = \frac{81}{2} = 40,5.\]

\[Ответ:\ \ 40,5.\]