Question

If three positive real numbers $\mathrm{a},\mathrm{b},\mathrm{c}$ are in A. P. and if $\mathrm{abc}=64$ then the minimum value of $\mathrm{b}$ is

• A. $6$
• B. $5$
• C. $4$
• D. $3$

Answer 1

The correct option is C

$4$

Explanation for the correct option:

Finding the minimum possible value of $\mathbf{b}$:

Since, $\mathrm{a},\mathrm{b},\mathrm{c}$ are in A.P then,

$2\mathrm{b}=\mathrm{a}+\mathrm{c}...\left(1\right)$

We know, $\mathrm{AM}\ge \mathrm{GM}$, for positive quantities, that

$$\begin{gathered} \left(\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{3}\right)\ge \sqrt[3]{\mathrm{abc}} \\ \Rightarrow \frac{2\mathrm{b}+\mathrm{b}}{3}\ge \sqrt[3]{64}\mathbf{[}\mathbf{\because }\mathbf{}\mathbf{from}\mathbf{}\mathbf{1}\mathbf{]} \\ \Rightarrow \frac{3\mathrm{b}}{3}\ge {\left(4^3\right)}^{\frac{1}{3}} \\ \Rightarrow \mathrm{b}\ge 4 \end{gathered}$$

Hence, the correct option is C.