Sequence and Series
AM-GM Inequality
JEE Main 2017
Grade 11
Question:
If three positive numbers $a, b, c$ are in A.P. such that $abc = 8$, then the minimum possible value of $b$ is:
(1) $1$
(2) $2$
(3) $3$
(4) $4$
Step-by-Step Solution
Key Concept: In A.P., $a + c = 2b$. By AM-GM: $b = (a + c)/2 \geq \sqrt{ac}$, so $b^2 \geq ac$. Since $abc = 8$ and $ac \leq b^2$, get $b^3 \geq b \cdot ac = 8$. Conclude.
The detailed step-by-step mathematical proof is available inside the Mathbee app workspace.
Correct Answer: (2)