Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 422

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

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

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

\[= \frac{1}{8} \bullet \frac{2}{3} = \frac{1}{4 \bullet 3} = \frac{1}{12}.\]

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

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

\[b_{1} = \frac{b_{5}}{q^{4}} = \frac{1}{81}\ :\left( \frac{1}{3} \right)^{4} =\]

\[= \frac{1}{81}\ :\frac{1}{81} = 1;\]

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

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

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

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

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

\[= 9 \bullet \frac{3}{4} = \frac{27}{4} = 6,75.\]

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

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

\[\ b_{1} = \frac{b_{4}}{q^{3}} = \frac{1}{8}\ :\left( - \frac{1}{2} \right)^{3} =\]

\[= \frac{1}{8}\ :\left( - \frac{1}{8} \right) = - 1;\]

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

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

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