Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 18

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\[1)\ q = - \frac{1}{2}\text{\ \ }и\ \ b_{1} = \frac{1}{8}\]

\[S = \frac{b_{1}}{1 - q} = \frac{\frac{1}{8}}{1 - \left( - \frac{1}{2} \right)} =\]

\[= \frac{1}{8}\ :\left( 1 + \frac{1}{2} \right) = \frac{1}{8}\ :\frac{3}{2} = \frac{1}{8} \bullet \frac{2}{3} =\]

\[= \frac{1}{4 \bullet 3} = \frac{1}{12}\]

\[Ответ:\ \ \frac{1}{12}.\]

\[2)\ q = \frac{1}{3}\text{\ \ }и\ \ b_{5} = \frac{1}{81}\]

\[b_{5} = b_{1} \bullet q^{4} \Longrightarrow b_{1} = \frac{b_{5}}{q^{4}} =\]

\[= \frac{1}{81}\ :\left( \frac{1}{3} \right)^{4} = \frac{1}{81}\ :\frac{1}{81} = 1\]

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

\[= 1 \bullet \frac{3}{2} = 1,5\]

\[Ответ:\ \ 1,5.\]

\[3)\ q = - \frac{1}{3}\text{\ \ }и\ \ b_{1} = 9\]

\[S = \frac{b_{1}}{1 - q} = \frac{9}{1 - \left( - \frac{1}{3} \right)} =\]

\[= 9\ :\left( 1 + \frac{1}{3} \right) = 9\ :\frac{4}{3} = 9 \bullet \frac{3}{4} =\]

\[= \frac{27}{4} = 6,75\]

\[Ответ:\ \ 6,75.\]

\[4)\ q = - \frac{1}{2}\text{\ \ }и\ \ b_{4} = \frac{1}{8};\]

\[b_{4} = b_{1} \bullet q^{3} \Longrightarrow \ b_{1} = \frac{b_{4}}{q^{3}} =\]

\[= \frac{1}{8}\ :\left( - \frac{1}{2} \right)^{3} = \frac{1}{8}\ :\left( - \frac{1}{8} \right) = - 1\]

\[S = \frac{b_{1}}{1 - q} = \frac{- 1}{1 - \left( - \frac{1}{2} \right)} =\]

\[= - 1\ :\left( 1 + \frac{1}{2} \right) = - 1\ :\frac{3}{2} =\]

\[= - 1 \bullet \frac{2}{3} = - \frac{2}{3}\]

\[Ответ:\ \ - \frac{2}{3}.\]