Question:
If three positive numbers a, b and c are in A.P. such that \(abc = 8\), then the minimum possible value of b is :
If three positive numbers a, b and c are in A.P. such that \(abc = 8\), then the minimum possible value of b is :
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Use a + c = 2b and the AM-GM inequality on a and c.
Updated On: Nov 11, 2026
- 2
- \(4^{1/3}\)
- \(4^{2/3}\)
- 4
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The Correct Option is A
Solution and Explanation
Since a, b, c are in A.P., \(a + c = 2b\), i.e. \(b = \frac{a+c}{2}\).
By AM-GM, \(\frac{a+c}{2} \ge \sqrt{ac}\), so \(ac \le b^2\).
\(abc = 8 \Rightarrow b \cdot ac = 8 \le b \cdot b^2 = b^3 \Rightarrow b \ge 2\).
Equality holds at \(a = b = c = 2\) (with \(abc = 8\)), so the minimum value of b is 2.
Answer: (1) 2
By AM-GM, \(\frac{a+c}{2} \ge \sqrt{ac}\), so \(ac \le b^2\).
\(abc = 8 \Rightarrow b \cdot ac = 8 \le b \cdot b^2 = b^3 \Rightarrow b \ge 2\).
Equality holds at \(a = b = c = 2\) (with \(abc = 8\)), so the minimum value of b is 2.
Answer: (1) 2
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