Найдите первый член геометрической прогрессии, в которой: q=2/3; S4=65.

\[q = \frac{2}{3};\ \ S_{4} = 65:\]

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

\[= \frac{b_{1}\left( \frac{16}{81} - 1 \right)}{- \frac{1}{3}} = - 3b_{1}\left( - \frac{65}{81} \right) =\]

\[= \frac{65b_{1}}{27}\]

\[\frac{65b_{1}}{27} = 65\]

\[b_{1} = \frac{65 \cdot 27}{65} = 27.\]