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(B) Let the three positive real numbers in $A$.$P$. be $(b-d)$, $b$, and $(b+d)$, where $d$ is the common difference.Given that their product is $4$,

Vedclass · Accepted Answer

$2^{2/3}$

Vedclass · Answer

(B) Let the three positive real numbers in $A$.$P$. be $(b-d)$, $b$, and $(b+d)$, where $d$ is the common difference.Given that their product is $4$, we have $(b-d)b(b+d) = 4$.This simplifies to $b(b^2 - d^2) = 4$, which implies $b^3 - bd^2 = 4$.Rearranging the terms, we get $b^3 = 4 + bd^2$.Since $b$ and $d^2$ are positive real numbers, $bd^2 \geq 0$.Therefore, $b^3 = 4 + bd^2 \geq 4$.Taking the cube root on both sides, we get $b \geq 4^{1/3} = (2^2)^{1/3} = 2^{2/3}$.Thus, the minimum possible value of $b$ is $2^{2/3}$.

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

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