Question

Let three positive real numbers $a,b,c$ are in $A.P.$ If $abc=4$ and the minimum value of $b$ is $2^k,$ then the value of $6k$ is

Answer 1

Let the common difference be $d$
$b=a+d\Rightarrow a=b-dc=b+d$
Now, $abc=4$
$\Rightarrow (b-d)b(b+d)=4\Rightarrow b(b^2-d^2)=4$
As $b^2\ge b^2-d^2$,
$\Rightarrow b\times b^2\ge b\times (b^2-d^2)\Rightarrow b^3\ge 4\Rightarrow b\ge 2^{2/3}$
Therefore, the minimum of $b$ is $2^{2/3}$.
Hence, $6k=4$