Question

Геометрическая прогрессия. Задание на фото.​

Answer 1

$\left \{ {{b_{1}+{b_{3}=-\frac{5}{8}} \atop {{b_{2}}+{b_{4}=\frac{5}{16}} \right.\\\\\left \{ {{{b_{1}+{b_{1}q^{2}=-\frac{5}{18}} \atop {{b_{1}q+{b_{1}q^{3}=\frac{5}{16}}} \right.\\\\:\left \{ {{{b_{1}q(1+q^{2})=\frac{5}{16}} \atop {{b_{1}(1+q^{2})=-\frac{5}{8}} \right.\\--------\\\boxed{q=-\frac{1}{2}}\\\\{b_{1}=-\frac{5}{8}:(1+q^{2})=-\frac{5}{8}:(1+\frac{1}{4})=-\frac{5}{8}:\frac{5}{4}=-\frac{5*4}{8*5}=-\frac{1}{2}$

$\boxed{b_{1} =-\frac{1}{2}} \\\\S_{6}=\frac{b_{1}*(q^{6}-1)}{q-1}=\frac{-\frac{1}{2}(\frac{1}{64}-1)}{-\frac{1}{2}-1 } =\frac{-\frac{1}{2}*(-\frac{63}{64})}{-\frac{3}{2}}=-\frac{21}{64}=-0,328125\\\\\boxed{S_{6}=-0,328125}$