## Precalculus (6th Edition) Blitzer

$a_2=-6$ and $a_3=18$
Here, we have $a_1=2$ and $a_2=a_1 r= 2r ;\\ a_3=a_1 r^2= 2r^2 ;\\ a_4=a_1 r^3= 2r^3=-54$ Now, $\dfrac{a_4}{a_1}=\dfrac{2r^3}{2}$ $r^3=-27 \implies r=-3$ Now, $a_2=a_1 r= 2(-3)=-6 ;\\ a_3=a_1 r^2= 2(-3)^2=18$ Hence, $a_2=-6$ and $a_3=18$