Find four consecutive even integers such that the sum of the second and fourth and integer is one half times the sum of the first and third integers

Okay lets create an eqn from that information

A first int, B second int, C third int, D fourth int.

B = A + 2
C = A + 4
D = A + 6
A is the smallest integer

B + D = 0.5 (A + C) 

Now lets substitute

(A + 2) + (A + 6) = 0.5(A + (A + 4))
now lets dist

2a + 8 = 0.5(2a +4)
2a + 8 = a + 2
a + 8 = 2
a = -6

B = -6 +2
B = -4
C = -6 + 4
C = -2
D = -6 + 6
D = 0

Now using B + D = 0.5(A + C)
-4 + 0 = 0.5(-6 + (-2))
-4 = 0.5 (-8)
-4 = -4

Correct 
Therefore, First integer is -6, second integer is -4, third integer is -2 and fourth integer is 0

72 is the mode because it shows up the most number of times