Найдите сумму бесконечной геометрической прогрессии (bn), если b2=36, b4=16.

Ответ:

\[b_{2} = 36;\ \ \ \ \ b_{4} = 16:\]

\[\left\{ \begin{matrix} b_{1}q = 36\ \\ b_{1}q^{3} = 16 \\ \end{matrix}\text{\ \ \ \ \ \ \ } \right.\ \]

\[\left\{ \begin{matrix} b_{1} = \frac{36}{q}\text{\ \ \ \ \ \ \ \ \ } \\ \frac{36}{q} \cdot q^{3} = 16 \\ \end{matrix}\text{\ \ \ \ \ \ \ } \right.\ \]

\[\left\{ \begin{matrix} b_{1} = \frac{36}{q} \\ q^{2} = \frac{16}{36} \\ \end{matrix}\text{\ \ \ \ \ \ \ \ } \right.\ \]

\[\left\{ \begin{matrix} b_{1} = \frac{36}{q} \\ q = \pm \frac{2}{3} \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} b_{1} = 54 \\ q = \frac{2}{3}\text{\ \ \ \ } \\ \end{matrix}\ \ \ \ \ \ или\ \ \ \ \right.\ \text{\ \ \ \ }\left\{ \begin{matrix} b_{1} = - 54 \\ q = - \frac{2}{3}\text{\ \ \ } \\ \end{matrix} \right.\ \]

\[S = \frac{54}{1 - \frac{2}{3}} = \frac{54}{\frac{1}{3}} = 162.\ \ \ \]

\[S = \frac{- 54}{1 + \frac{2}{3}} = \frac{- 54}{\frac{5}{3}} =\]

\[= \frac{- 54 \cdot 3}{5} = - 32,4.\]

\[Ответ:162\ \ или\ - 32,4.\]