[tex]a_1=4;\ a_2=-12\\\\r=\dfrac{a_2}{a_1}\to r=\dfrac{-12}{4}=-3[/tex]
The formula of a geometric sequence:
[tex]a_n=a_1\cdot r^{n-1}[/tex]
substitute
[tex]a_n=4\cdot(-3)^{n-1}[/tex]
[tex]The\ domain:\ \text{all integers where}\ n\geq1[/tex]
Answer: [tex]\boxed{a_n=4(-3)^{n-1};\ \text{all integers where}\ n\geq1}[/tex]