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JEE Main 2017 · 9 Apr · Shift 1, Mathematics question 68

If three positive numbers a, b and c are in A.P. such that abc = 8, then the minimum possible value of b is :

Choose one option
Answer and worked solution

Official answer: A

b=a+c2≥ac=8bb=\frac{a+c}2\ge\sqrt{ac}=\sqrt{\frac8b}, so b3≥8b^3\ge8 and b≥2b\ge2, with equality at a=b=c=2a=b=c=2.

Topics:AM-GM-HM inequalitiesArithmetic progression
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