Даны четыре первых члена геометрической прогрессии. Сумма двух крайних членов равна 52, а двух средних равна 16. Найдите эти члены.

Ответ:

\[b_{1};b_{2};b_{3};b_{4};\ldots\]

\[b_{1} + b_{4} = 52\]

\[b_{1} + b_{1}q^{3} = 52\]

\[b_{1}\left( 1 + q^{3} \right) = 52\]

\[b_{2} + b_{3} = 16\]

\[b_{1}q + b_{1}q^{2} = 16\]

\[b_{1}\left( q + q^{2} \right) = 16.\]

\[\frac{1 + q^{3}}{q + q^{2}} = \frac{52}{16}\]

\[4 \cdot \left( 1 + q^{3} \right) = 13 \cdot \left( q + q^{2} \right)\]

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

\[= 13q(1 + q)\]

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

\[4q^{2} - 4q - 13q + 4 = 0\]

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

\[D = 289 - 64 = 225\]

\[q_{1} = \frac{17 + 15}{8} = \frac{32}{8} = 4;\ \ \ \]

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

\[При\ q = 4:\]

\[b_{1} = \frac{52}{1 + 64} = \frac{52}{65} = \frac{4}{5};\]

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

\[b_{3} = \frac{16}{5} \cdot 4 = \frac{64}{5};\]

\[b_{4} = \frac{64}{5} \cdot 4 = \frac{256}{5}.\]

\[При\ q = \frac{1}{4}:\]

\[b_{1} = \frac{52}{1 + \frac{1}{64}} = \frac{52}{\frac{65}{64}} = \frac{52 \cdot 64}{65} =\]

\[= \frac{256}{5};\]

\[b_{2} = \frac{256}{5} \cdot \frac{1}{4} = \frac{64}{5};\]

\[b_{3} = \frac{64}{5} \cdot \frac{1}{4} = \frac{16}{5};\]

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