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, y = y, z = y + d,

where d is the common difference.

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

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

Now, we can use the AM-GM inequality, which states that the arithmetic mean of a set of non-negative numbers is always greater than or equal to the geometric mean. Applying this to y, y, and 27 + d^2 * y, we get:

(y + y + 27 + d^2 * y) / 3 ≥ ((y * y * (27 + d^2 * y)) ^ (1/3)), (2y + 27 + d^2 * y) / 3 ≥ (27 + d^2 * y) ^ (1/3), (2y + 27 + d^2 * y) ≥ 3 * (27 + d^2 * y) ^ (1/3).

Since we want to find the minimum value of y, we need to minimize the right-hand side. The minimum value of (27 + d^2 * y) ^ (1/3) is achieved when d = 0 (because d^2 * y is always non-negative and we want to minimize the expression). Substituting d = 0, we get:

2y + 27 ≥ 3 * 27 ^ (1/3), 2y ≥ 3 * 3 - 27, 2y ≥ 9 - 27, 2y ≥ -18, y ≥ -9.

So, the minimum possible value of y is -9, but since y is a positive number, the minimum possible value of y is 0. However, this is not in the options. This is because in the context of the problem, y cannot be 0 (since xyz = 27 and if y = 0, then xyz = 0). Therefore, the minimum possible value of y from the given options is 1.3 (Option A).