[tex]\displaystyle\bf\\\left \{ {{b_{2} =-4} \atop {b_{1}+b_{3} =10 }} \right.\\\\\\\left \{ {{b_{1} \cdot q=-4} \atop {b_{1}+b_{1}\cdot q^{2} =10 }} \right.\\\\\\\left \{ {{b_{1} \cdot q=-4} \atop {b_{1}\cdot(1+ q^{2}) =10 }} \right\\\\\\\left \{ {{b_{1} =-\dfrac{4}{q} } \atop {b_{1}\cdot(1+ q^{2}) =10 }} \right\\\\\\-\frac{4}{q} \cdot(1+q^{2})=10 \ \ ; \ \ q\neq 0\\\\\\-4-4q^{2} -10q=0\\\\2q^{2} +5q+2=0\\\\D=5^{2} -4\cdot 2\cdot 2=25-16=9=3^{2} \\\\\\q_{1}=\frac{-5-3}{4}=-2[/tex]

[tex]\displaystyle\bf\\q_{2}=\frac{-5+3}{4}=-0,5 \\\\\\b_{1}' =-4:q_{1} =-4:(-2)=2\\\\b_{1}''= -4:(-0,5)=8\\\\\\b_{6}'=b_{1}'\cdot q_{1} ^{5} =2\cdot (-2)^{5} =2\cdot(-32)=-64\\\\b_{6}''=b_{1}''\cdot q_{2} ^{5} =8\cdot (-0,5)^{5} =8\cdot(-\frac{1}{32} )=-0,25[/tex]