Progressions - 1 (IIFT-2008)

If the positive real numbers a, b and c are in Arithmetic Progression, such that abc = 4, then minimum possible value of b is:

(A)  2^3/2

(B)   2^2/3

(C)   2^1/3

(D)None of the above
Solution follows here:

Solution:

Given that a,b,c are in A.P. ∴ b = (a+c)/2                      ---- (1)

And also given that,     abc =4 => b = 4/ac                  ---- (2)

From (1) and (2),

(a+c)/2 = 4/ac   => (a+c)ac = 8           ---- (3)

For b to be minimum in A.P, a=c             ---- (4)

From (3) and (4),

(2a)a² = 8 => a³ = 4 => a = 4^1/3                      

From (1) and (4), b = 2a/2 = a = 4^1/3  = 2^2/3

Answer(B)