Question

Члены арифметической прогрессии удовлетворяют условию а3+а6=-20, а s6=-72. тогда а11

Answer 1

$\left \{ {{a_3+a_6=-20} \atop {\frac{a_1+a_6}{2}*6=-72}} \right. <= \left \{ {{a_1+2d+a_1+5d=-20} \atop {a_1+a_1+5d=-24}} \right. <=$
$<= \left \{ {{2a_1+7d=-20} \atop {2a_1+5d=-24}} \right. <= \left \{ {{2d=4} \atop {2a_1=-24-5d}} \right. <=$
$<= \left \{ {{d=2} \atop {a_1=-17}} \right. =a_{11}=a_1+10d=-17+10*2=3$
ответ: 3.

Answer 2

A3+a6=-20
a1+2d+a1+5d=-20
2a1=-20-7d
S6=(2a1+5d)*6/2=-72
2a1+5d=-72:3=-24
-20-7d+5d=-24
-2d=-24+20=-4
d=-4:(-2)=2
a1=(-20-7d)/2=(-20-14)/2=-34/2=-17
a11=a1+10d
a11=-17+10*2=-17+20=3