Rule: divide by 2, then add 6. 56, 100, 34, 23 a. 23, 34, 56, 100 b. 23, 56, 34, 100 c. 100, 34, 56, 23 d. 100, 56, 34, 23

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.