## Intermediate Algebra for College Students (7th Edition)

$a_2=12$ and $a_3=18$
Given: Geometric sequence: $8,a_2,a_3,27$ Here $a_4=a_1r^{n-1}=a_1r^3$ Thus, $27=8r^3 \implies r=\dfrac{3}{2}$ Now, $a_2=a_1r^{2-1}=a_1r \implies (8)(\dfrac{3}{2})=12$ and $a_3=a_1r^{3-1}=a_1r^2 \implies (12)(\dfrac{3}{2})=18$