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

Since a,b, c are in A.P., therefore, b -a = d and c -b = d,where dis the common difference of the A.P.∴ a=b−d and c=b+dNow, abc=4⇒ (b−d)b(b+d)=4⇒ bb2−d2=4But, bb2−d24⇒ b>22/3Hence, the minimum value of b is  2 2/3