Найдите сумму первых четырёх членов геометрической прогрессии (bn), в которой: b3=1/25, b4=1/125.

\[b_{3} = \frac{1}{25};\ \ \ b_{4} = \frac{1}{125}:\]

\[q = \frac{b_{4}}{b_{3}} = \frac{1}{125} \cdot 25 = \frac{1}{5};\]

\[b_{3} = b_{1} \cdot q^{2}\]

\[b_{1} = \frac{b_{3}}{q^{2}} = \frac{1}{25}\ :\left( \frac{1}{5} \right)^{2} = \frac{1}{25} \cdot 25 =\]

\[= 1.\]

\[S_{4} = \frac{1 \cdot \left( \left( \frac{1}{5} \right)^{4} - 1 \right)}{\frac{1}{5} - 1} =\]

\[= \frac{\frac{1}{625} - 1}{- \frac{4}{5}} = \frac{- \frac{624}{625}}{- \frac{4}{5}} = \frac{624 \cdot 5}{625 \cdot 4} =\]

\[= \frac{156}{125} = 1\frac{31}{125}.\]