Q. If three positive real numbers $a,\ b,\ c$ are in AP and $abc = 4$, then the minimum possible value of $b$ is
A) $2^{\frac{3}{2}}$
B) $2^{\frac{2}{3}}$
C) $2^{\frac{1}{3}}$
D) $2^{\frac{5}{2}}$
Explanation
(b)
Since, $a,b\$and$\ c$ are in AP.
Let $d$ be the common difference.
$\therefore$ $a = b - d,\ b = d,\ c = b + d$
Also, $abc = 4$
$\Rightarrow \ (b - d)d(b + d) = 4$
$\Rightarrow \left( b^{2} - d^{2} \right)b = 4$
$\Rightarrow b^{3} = 4 + d^{2}b$
$\Rightarrow \ b^{3} \geq 4$
$\Rightarrow b \geq (2)^{\frac{2}{3}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.