For this problem, it is described that a set of number (x, y, z, ...) is calculated by a series of identical operation using the preceding number. In this problem if we let x as the first number, the second number y is, y = (x/2) + 6. Then the 3rd number z is equal to, z = (y/2) + 6, so on. Thus for this problem, the solution is D: 100, 56, 34, 23 through trial and error.
