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

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

\[S_{4} = b_{1} \cdot \frac{q^{4} - 1}{q - 1}\]

\[65 = b_{1} \cdot \frac{\frac{16}{81} - 1}{\frac{2}{3} - 1}\]

\[65 = b_{1} \cdot \frac{\frac{65}{81}}{\frac{1}{3}}\]

\[65 = b_{1} \cdot \frac{65 \cdot 3}{81}\]

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

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

\[Ответ:27.\]