With $$a_3=a_1+2d,a_4=a_1+3d,a_5=a_1+4d,a_7=a_1+6d$$ we get the system $$3a_1+12d=-12$$ and
$$(a_1+2d)(a_1+3d)(a_1+4d)=-90$$ Can you solve this=
With $$a_3=a_1+2d,a_4=a_1+3d,a_5=a_1+4d,a_7=a_1+6d$$ we get the system $$3a_1+12d=-12$$ and
$$(a_1+2d)(a_1+3d)(a_1+4d)=-90$$ Can you solve this=