Question

If three positive real numbers a, b, c are in A.P., with $abc=4$, then the minimum value of b is-

• A. $4^{1/3}$
• B. 3
• C. 2
• D. 1/2

Answer 1

The correct option is A $4^{1/3}$

Let $d$ be the common difference of the A.P.

Given, $4=abc$

$=(b-d)b(b+d)$

$=b(b^2-d^2)$

$=b^3-b{d}^2$

Therefore, $b^3=4+b{d}^2$.

Since $b>0$ and $d^2\ge 0$, so $b^3\ge 4$.

$\Rightarrow b\ge 4^{\frac{1}{3}}$