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

Answer 1

The correct option is B $2^{2/3}$

Since $a,b,c$ are in A.P., therefore, $b-a=d$ and $c-b=d$, where d is the common difference of the A.P.
$\therefore a=b-d$ and $c=b+d$
Now,
$abc=4$
$\Rightarrow (b-d)b(b+d)=4$
$\Rightarrow b(b^2-d^2)=4$
But,
$b(b^2-d^2)<b\times b^2$
$\Rightarrow b(b^2-d^2)<b^3$
$\Rightarrow 4<b^3$
$\Rightarrow b^3>4$
$\Rightarrow b>2^{2/3}$
Hence, the minimum value of $b$ is $2^{2/3}$.