Question

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

• A. $2^{\frac{1}{3}}$
• B. $2^{\frac{2}{3}}$
• C. $2^{\frac{1}{2}}$
• D. $2^{\frac{3}{2}}$

Answer 1

The correct option is B

$2^{\frac{2}{3}}$

Since a, b, c are in A.P., therfore, $b-a=d$ and $c-b=d$, where d is the common difference of the A.P.

$\therefore a=b-dandc=b+d$

Now,

$abc=4$

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

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

But,

$b(b^2-d^2)<b\times b^2$

$\Rightarrow b(b^2-d^2)<b^3$

$\Rightarrow 4<b^3$

$\Rightarrow b^3>4$

$\Rightarrow b>2^{\frac{2}{3}}$

Hence, the minimun value of b is $2^{\frac{2}{3}}$