If three positive real numbers `a,b ,c` are in A.P such that `a b c=4` , then the minimum value of `b` is a)`2^(1//3)` b) `2^(2//3)` c) `2^(1//2)` d) `2^(3//23)`

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 AP. Therefore,
b-a=d and c-b=d, where d is the common difference of the AP.
`rArr" "a=b-d and c=b+d` Now,
abc=4
`rArr" "(b-d)b(b+d)=4`
`rArr" "b(b^(2)-d^(2))=4`
But, `b(b^(2)-d^(2))ltb.b^(2)" "[because" "b^(2)-d^(2)ltb^(2)]`
`rArr" "b(b^(2)-d^(2))ltb^(3)`
`rArr" "4ltb^(3)rArrb^(3)gt4rArrbgt2^(2//3)`
Hence, the minimum value of b is `2^(2//3)`.