Resposta :
[tex](a_1,a_2,a_3,\dots,a_{12},\dots)=(-15,-9,-3,\dots)[/tex]
[tex]\boxed{a_n=a_k+(n-k)r}\ \to\ a_2=a_1+r\ \therefore\ -9=-15+r\ \therefore\ \boxed{r=6}[/tex]
[tex]a_{12}=a_1+11r\ \therefore\ a_{12}=-15+11(6)=-15+66\ \therefore\ \boxed{a_{12}=51}[/tex]