Q 56. If three positive numbers x, y and z are in A.P. such that xyz = 27 then the minimum possible value of y is: Ops: A. 1.3 B. 3 C. 6 D. 6.3

This problem can be solved using the Arithmetic Progression (AP) and AM-GM inequality concepts.

Given that x, y, and z are in AP, we can write them as:

x = y - d, z = y + d,

where d is the common difference.

Also, given that xyz = 27, we can substitute x and z from above:

(y - d)(y)(y + d) = 27, y^3 - d^2*y = 27.

Now, we need to minimize y. This can be done using the AM-GM inequality, which states that the arithmetic mean of a set of positive numbers is always greater than or equal to the geometric mean.

Applying this to y - d, y, and y + d, we get:

[(y - d) + y + (y + d)] / 3 ≥ ((y - d)y(y + d))^(1/3), 3y / 3 ≥ (y^3 - d^2y)^(1/3), y ≥ (y^3 - d^2y)^(1/3).

Since y^3 - d^2*y = 27, we get:

y ≥ 27^(1/3), y ≥ 3.

So, the minimum possible value of y is 3. Therefore, the correct option is B. 3.