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

For three positive numbers, We have AP ge GP (a + b + c / 3) ge (abc)^1/3 Since a, b, c are in A.P., we have\ (a + b + c / 3) = b \ => b ge (abc)^1/3 Given that abc = 8, b ge 2 Hence, the minimum possible value of b is boxed2

If three positive numbers a,b and c are in A.P. such th... | Sequence & Series (Mathematics)

For three positive numbers, $\\$ We have $AP \ge {GP}$ $\frac{a + b + c}{3} \ge (abc)^{1/3}\\$ Since $a, b, c$ are in A.P., we have $\\ \frac{a + b + c}{3} = b$ $\\ \Rightarrow b \ge (abc)^{1/3}\\$ Given that $abc = 8$ , $b \ge 2\\$ Hence, the minimum possible value of $b$ is $\boxed{2}$

Explanation

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