Chapter 10 - Section 10.3 - Geoetric Sequences and Series - Exercise Set - Page 1074: 63

$ a_2=12$ and $ a_3=18 $

Here, we have $ a_1=8$ and $ a_2=a_1 r= 8r ;\\ a_3=a_1 r^2= 8r^2 ;\\ a_4=a_1 r^3= 8r^3=27$ Now, $\dfrac{a_4}{a_1}=\dfrac{8r^3}{8}$ $ r^3=\dfrac{27}{8} \implies r=\dfrac{3}{2}$ Now, $ a_2=a_1 r= 8(\dfrac{3}{2})=12 ;\\ a_3=a_1 r^2= 8(\dfrac{3}{2})^2=18 $ Hence, $ a_2=12$ and $ a_3=18 $