1)Дано 2) Дано 3) Дано (bn) (bn) (bn) b1= -125 b5=27 b2=0,08 q= 0,2 q=корень из 3 b5=0,64...

Решите задачу:

$1)b_{1}=-125\\\\q=0,2\\\\b_{5}=b_{1}*q^{4}=-125*0,2^{4}=-5^{3}*\frac{1}{5^{4} }=-\frac{1}{5}=-0,2$

$2)b_{5}=27\\\\q=\sqrt{3}\\\\b_{5}=b_{1}*q^{4} \\\\b_{1}=\frac{b_{5} }{q^{4} }=\frac{27}{(\sqrt{3} )^{4} }=\frac{27}{9}=3$

$3)\left \{ {{b_{5}=0,64 } \atop {b_{2}=0,08 }} \right.\\\\:\left \{ {{b_{1}*q^{4}=0,64} \atop {b_{1}*q=0,08}} \right.\\ ------\\q^{3}=8\\\\q=\sqrt[3]{8}=2\\\\b_{1}=\frac{0,08}{2} =0,04\\\\S_{9} =\frac{b_{1}(q^{9}-1) }{q-1}=\frac{0,04(2^{9} -1)}{2-1}=0,04(512-1)=0,04*511=20,44$