(b) Since, $a,\ b,c$ are in AP. $\Rightarrow 2b + a + c$ $\therefore\ 3^{2b} = 3^{a + c} \Rightarrow \left( 3^{b} \right)^{2} = 3^{a} \bullet 3^{c}$ $ie,\ 3^{a},\ 3^{b},\ 3^{c}$ are in GP.
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.