use am- gm ineuqality .
a+b+c/3 ≥(abc)1/3
and 2b=a+c
If three positive real numbers a, b, c are in A.P. and abc = 4 then minimum possible value of b is
(b-d)b(b+d)=4
→b3-bd2=4
→b3=4+bd2
b is least when 4+bd2 is least
as all are +ve
least bd2 is 0
hence least value of b= 4^(1/3)
..........HEY d is common difference