Найти сумму семи первых членов бесконечной геометрической прогрессии со знаменателем |q|...

$S_{7}= \frac{b_{1}}{1-q}$
$b_{2}=4$
$\frac{b^{2}_{1}+b^{2}_{2}+...+b^{2}_{7}}{b_{1}+b_{2}+...+b_{7}}= \frac{16}{3}$
$\frac{b^{2}_{1}+b^{2}_{2}+...+b^{2}_{7}}{S_{7}}= \frac{16}{3}$

$b_{2}=b_{1}q$ => $b_{1}= \frac{b_{2}}{q}= \frac{4}{q}$
$S_{7}=S= \frac{4}{q(1-q)}$

$\frac{b^{2}_{1}*(1+q^{2}+q^{4}+...+q^{12})}{ \frac{4}{q(1-q)} }= \frac{16}{3}$
$\frac{ \frac{16}{q^{2}} *(1+ \frac{q^{2}}{1-q^{2}} )}{ \frac{4}{q(1-q)} }= \frac{16}{3}$
$\frac{16q*(1-q)}{q^{2}}* \frac{1-q^{2}+q^{2}}{1-q^{2}}= \frac{16*4}{3}$
$\frac{16}{q(1+q)}= \frac{16*4}{3}$
$\frac{1}{q(1+q)}= \frac{4}{3}$
$q+q^{2}- \frac{3}{4}=0$
$4q^{2}+4q-3=0, D=64$
$q_{1}= \frac{-4-8}{8}=- \frac{12}{8}\ \textless \ -1$
$q_{2}= \frac{1}{2}$

$S=\frac{4}{0.5*(1-0.5)}=\frac{4}{0.5*0.5}=16$