[tex]\displaystyle\bf\\-\left \{ {{b_{1} + b_{2} + b_{3} =6} \atop {b_{2} + b_{} + b_{4} =-3}} \right. \\-----------\\b_{1} + b_{2} + b_{3} -b_{2} - b_{3} - b_{4} =6-(-3)\\\\b_{1} -b_{4} =9\\\\b_{1} -b_{1} \cdot q^{3} =9\\\\\boxed{b_{1} \cdot(1-q^{3} )=9}\\\\\\S_{3} =b_{1} + b_{2} + b_{3} =6\\\\S_{3} =6\\\\S_{3}=\frac{b_{1} \cdot(1-q^{3} )}{1-q} \\\\\\6=\frac{9}{1-q} \\\\9=6\cdot(1-q)\\\\9=6-6q\\\\6q=-3\\\\q=-0,5[/tex]
[tex]\displaystyle\bf\\b_{1} =\frac{9}{1-q^{3} } =\frac{9}{1-(-0,5)^{3} } =\frac{9}{1+0,125} =\frac{9}{1,125} =8\\\\\\b_{2} =b_{1} \cdot q=8\cdot(-0,5)=-4\\\\b_{3} =b_{2} \cdot q=-4\cdot(-0,5)=2\\\\b_{4} =b_{3} \cdot q=2\cdot(-0,5)=-1\\\\\\Otvet \ : \ 8 \ ; \ -4 \ ; \ 2 \ ; \ -1[/tex]
Copyright © 2024 SCHOLAR.TIPS - All rights reserved.
Answers & Comments
[tex]\displaystyle\bf\\-\left \{ {{b_{1} + b_{2} + b_{3} =6} \atop {b_{2} + b_{} + b_{4} =-3}} \right. \\-----------\\b_{1} + b_{2} + b_{3} -b_{2} - b_{3} - b_{4} =6-(-3)\\\\b_{1} -b_{4} =9\\\\b_{1} -b_{1} \cdot q^{3} =9\\\\\boxed{b_{1} \cdot(1-q^{3} )=9}\\\\\\S_{3} =b_{1} + b_{2} + b_{3} =6\\\\S_{3} =6\\\\S_{3}=\frac{b_{1} \cdot(1-q^{3} )}{1-q} \\\\\\6=\frac{9}{1-q} \\\\9=6\cdot(1-q)\\\\9=6-6q\\\\6q=-3\\\\q=-0,5[/tex]
[tex]\displaystyle\bf\\b_{1} =\frac{9}{1-q^{3} } =\frac{9}{1-(-0,5)^{3} } =\frac{9}{1+0,125} =\frac{9}{1,125} =8\\\\\\b_{2} =b_{1} \cdot q=8\cdot(-0,5)=-4\\\\b_{3} =b_{2} \cdot q=-4\cdot(-0,5)=2\\\\b_{4} =b_{3} \cdot q=2\cdot(-0,5)=-1\\\\\\Otvet \ : \ 8 \ ; \ -4 \ ; \ 2 \ ; \ -1[/tex]