Question

If three positive real numbers $a, b, c$ are in $AP$ and $abc = 4$, then the minimum possible value of $b$ is

• A. $2^{3/2}$
• B. $2^{2/3}$
• C. $2^{1/3}$
• D. $2^{5/2}$

Answer 1

Answer: (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, $abc =4$

$\Rightarrow \left(b-d\right)d\left(b+d\right)=4$

$\Rightarrow \left(b^{2}-d^{2}\right)b=4$

$\Rightarrow b^{3}=4+d^{2}b$

$\Rightarrow b^{3}\ge\,4$

$\Rightarrow b \ge\, \left(2\right)^{2 /3}$