Question

знайдіть суму перших семи членів арефметичної прогресії, добуток третього й п'ятого членів якої дорівнюють другому члену, а сума першого і восьмого ленів дорівнює 2​

Answer 1

$\displaystyle\bf\\\left \{ {{a_{3}\cdot a_{5} =a_{2} } \atop {a_{1}+a_{8} =2}} \right. \\\\\\\left \{ {{(a_{1} +2d)\cdot(}{a_{1} +4d)} =a_{1} +d} \atop {a_{1}+a_{1}+7d=2 }} \right. \\\\\\\left \{ {{2a_{1} +7d=2} \atop {(a_{1} +2d)\cdot(}{a_{1} +4d)} =a_{1} +d}} \right. \\\\\\\left \{ {{a_{1} +3,5d=1} \atop {(a_{1} +2d)\cdot(}{a_{1} +4d)} =a_{1} +d}} \right. \\\\\\\left \{ {{a_{1} =1-3,5d} \atop {(1-3,5d +2d)\cdot(}{1-3,5d +4d)} =1-3,5d+d}} \right.$

$\displaystyle\bf\\\left \{ {{a_{1} =1-3,5d} \atop {(1-1,5d )\cdot(}{1+0,5d )} =1-2,5d}} \right. \\\\\\\left \{ {{a_{1}=1-3,5d } \atop {1+0,5d-1,5d-0,75d^{2} -1+2,5d=0}} \right. \\\\\\\left \{ {{a_{1} =1-3,5d} \atop {0,75d^{2} -1,5d =0}} \right. \\\\\\\left \{ {{a_{1} =1-3,5d} \atop {0,75d(d-2)=0}} \right.\\\\\\d_{1} =0-neyd \ \ ; \ \ d_{2} =2\\\\a_{1} =1-3,5\cdot 2=1-7=-6\\\\a_{7} =a_{1}+6d=-6+6\cdot 2=6\\\\\\S_{7} =\frac{a_{1}+a_{7} }{2} \cdot 7=\frac{-6+6}{2} \cdot 7=0$

$\displaystyle\bf\\Otvet:S_{7} =0$