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

Let d be the common difference of the AP. then 4=abc=(b−d)b(b+d)=b(b2−d2)⇒b3=4+bd2≥4 [∵b>0, d2≥0]⇒b≥22/3 Thus, minimum possible value of b is 22/3  , that is the case when d = 0.