Question 1096413
If the first term is {{{b}}} , and the common ratio is {{{r}}} ,
the second, third, and fourth terms are respectively
{{{br}}} , {{{br^2}}} and {{{br^3}}} .
The problem says that
{{{br+br^2=6}}} --> {{{br(1+r)=6}}} , and that
{{{br^2+br^3=-12}}} --->{{{br^2(1+r)=-12}}}
{{{system(br(1+r)=-6,br^2(1+r)=-12)}}} --> {{{br^2(1+r)/(br(1+r))=(-12)/6}}} -->{{{highlight(r=-2)}}}
{{{system(r=-2,br(1+r)=6)}}} --> {{{-2b(-2+1)=6}}} --> {{{-2b(-1)=6}}} --> {{{2b=6}}} --> {{{b=6/2}}} --> {{{highlight(b=3)}}}