Question

помогите решить геометрическую прогрессию​

Answer 1

$\left \{ {b_{5}=\frac{1}{3}} \atop {b_{3}-b_{1}=-24}} \right.\\\\\left \{ {{b_{1}*q^{4} =\frac{1}{3}} \atop {b_{1}q^{2}-b_{1}=-24}} \right.\\\\:\left \{ {{b_{1}*q^{4}=\frac{1}{3}} \atop {b_{1}(q^{2}-1)=-24}} \right.\\--------\\ \frac{q^{4} }{q^{2} -1}=-\frac{1}{72}\\\\72q^{4}=-q^{2}+1\\\\72q^{4}+q^{2}-1=0\\\\q^{2}=m,m>0\\\\72m^{2}+m-1=0\\\\D=1^{2}-4*72*(-1)=1+288=289=17^{2}\\\\m_{1}=\frac{-1+17}{144}=\frac{16}{144}\\\\m_{2}=\frac{-1-17}{144}=-\frac{18}{144} <0$

m₂ - не подходит

$q^{2}=\frac{16}{144}=\frac{1}{9} \\\\b_{1}=\frac{-24}{q^{2}-1 }=\frac{-24}{\frac{1}{9}-1}=\frac{-24}{-\frac{8}{9}} =27\\\\Otvet:\boxed{b_{1}=27 }$