Question

помогите, пожалуйста:(

Answer 1

$\left \{ {{b_{6}-b_{4}=27} \atop {b_{3}-b_{5}=9}} \right\\\\\left \{ {{b_{1}q^{5}-b_{1}q^{3}=27} \atop {b_{1}q^{2} -b_{1}q^{4}=9}} \right.\\\\:\left \{ {{b_{1}q^{3}(q^{2}-1)=27} \atop {b_{1}q^{2}(1-q^{2})=9}} \right.\\ --------\\\oxed{q=-3}\\\\b_{1}=\frac{9}{q^{2}(1-q^{2})}=\frac{9}{9(1-9)}=-\frac{1}{8}\\\\\boxed{b_{1}=-\frac{1}{8}}\\\\S_{3}=\frac{b_{1}(q^{3} -1)}{q-1}=\frac{-\frac{1}{8}*(-27-1) }{-3-1}=\frac{-\frac{1}{8}*(-28) }{-4}=-\frac{7}{8}=\boxed{-0,875}$