Даны первые четыре члена геометрической прогрессии. Сумма двух крайних членов равна 13, а двух средних равна 4. Найдите эти члены.

Ответ:

\[\left\{ \begin{matrix} b_{1} + b_{4} = 13 \\ b_{2} + b_{3} = 4\ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\left\{ \begin{matrix} b_{1} + b_{1}q^{3} = 13 \\ b_{1}q + b_{1}q^{2} = 4 \\ \end{matrix} \right.\ \]

\[\frac{1 + q^{3}}{q + q^{2}} = \frac{13}{4}\]

\[\frac{(1 + q)\left( 1 - q + q^{2} \right)}{q(1 + q)} = \frac{13}{4}\]

\[\frac{1 - q + q^{2}}{q} = \frac{13}{4}\]

\[4 - 4q + 4q^{2} = 13q\]

\[4q^{2} - 17q + 4 = 0\]

\[D = 289 - 64 = 225\]

\[q_{1} = \frac{17 + 15}{8} = 4;\ \ \]

\[q_{2} = \frac{17 - 15}{8} = \frac{1}{4}.\]

\[При\ q = 4:\]

\[b_{1} = \frac{13}{1 + q^{3}} = \frac{13}{1 + 4^{3}} = \frac{13}{65} = \frac{1}{5};\]

\[b_{2} = \frac{1}{5} \cdot 4 = \frac{4}{5};\]

\[b_{3} = \frac{4}{5} \cdot 4 = \frac{16}{5};\]

\[b_{4} = \frac{16}{5} \cdot 4 = \frac{64}{5}.\]

\[При\ q = \frac{1}{4}:\ \]

\[b_{1} = \frac{13}{1 + \left( \frac{1}{4} \right)^{3}} = \frac{13}{1 + \frac{1}{64}} =\]

\[= 13\ :\frac{65}{64} = 13 \cdot \frac{64}{65} = \frac{64}{5};\]

\[b_{2} = \frac{64}{5} \cdot \frac{1}{4} = \frac{16}{5};\]

\[b_{3} = \frac{16}{5} \cdot \frac{1}{4} = \frac{4}{5};\]

\[b_{4} = \frac{4}{5} \cdot \frac{1}{4} = \frac{1}{5}.\]