Question

очень нужно, помогите пожалуйста отдаю последние баллы

Answer 1

$\displaystyle\bf\\2) \ \left \{ {{b_{6}=-18 } \atop {b_{4} =-2}} \right. \\\\\\:\left \{ {{b_{1}\cdot q^{5} =-18 } \atop {b_{1} \cdot q^{3} =-2}} \right. \\---------\\q^{2} =9\\\\\\q_{1} =-3 \ \ \ ; \ \ \ q_{2}=3\\\\\\b'_{1} =\frac{-2}{q_{1}^{3} } =\frac{-2}{(-3)^{3} } =\frac{2}{27} \\\\\\b''_{1} =\frac{-2}{q_{2}^{3} } =\frac{-2}{3^{3} } =-\frac{2}{27}$

$\displaystyle\bf\\3) \ \left \{ {{b_{7}=16 } \atop {b_{2} =\dfrac{1}{2} }} \right. \\\\\\:\left \{ {{b_{1}\cdot q^{6} =16 } \atop {b_{1} \cdot q =\dfrac{1}{2} }} \right. \\---------\\q^{5} =32\\\\\\q=\sqrt[5]{32} =2\\\\\\b_{1} =\frac{1}{2}:q=\frac{1}{2}:2=\frac{1}{4}$