Сумма пятого и третьего членов геометрической прогрессии равна 90,а сумма второго и...

$\left \{ {{b_3+b_5}=90 \atop {b_2+b_4=-30}} \right.$
$\left \{ {{b_1q^2+b_1q^4}=90 \atop {b_1q+b_1q^3=-30}} \right.$
$\left \{ {{b_1q^2(1+q^2)}=90 \atop {b_1q(1+q^2)=-30}} \right.$
$\left \{ {{b_1(1+q^2)}= \frac{90}{q^2} \atop {q* \frac{90}{q^2} =-30}} \right.$
$\left \{ {{b_1(1+q^2)}= \frac{90}{q^2} \atop { \frac{90}{q} =-30}} \right.$
$\left \{ {{b_1(1+q^2)}= \frac{90}{q^2} \atop {}{q} =-3}} \right.$
$\left \{ {{b_1(1+(-3)^2)}= \frac{90}{9} \atop {}{q} =-3}} \right.$
$\left \{ {{b_1*10}= 10 \atop {}{q} =-3}} \right.$
$\left \{ {{b_1}= 1 \atop {}{q} =-3}} \right.$

$S_6= \frac{b_1(1-q^6)}{1-q}$
$S_6= \frac{1*(1-(-3)^6)}{1-(-3)} = \frac{1-729}{4} =-182$

Ответ: -182