If three positive real numbers a,b,c are in AP such that abc=4, then the minimum value of b is

Options

A. `2^(1//3)`
B. `2^(2//3)`
C. `2^(1//2)`
D. `2^(3//2)`

Answer

B

Explanation

Since a,b,c are in A.P., therefore, b-a=d and c-b=d, where d is the common difference of the A.P.
`therefore a=b-d and c=b+d`
Now,
abc=4
`rArr(b-d)b(b+d)=4`
`rArrb(b^(2)-d^(2))=4`
But, `b(b^(2)-d^(2))ltbxxb^(2)`
`rArrb(b^(2)-d^(2))ltb^(3)`
`rArr4ltb^(3)`
`rArrb^(3)gt4`
`rArrbgt2^(2//3)`
Hence, the minimum value of b is `2^(2//3)`.