Question

find four consecutive even integer such that the sum of the first three exceed the fourth by 8

Answer 1

Let the four consecutive terms in A.P are a+d, a+2d, a+3d, a+4d
From given condition
(a+d) + (a+2D) + (a+3d) = a+4d +8 
a+2+a+2x2 + a+ 3x2 = a x 4x2 + 8 { even number differ by 2)
3a + 12 = a + 16
3a -a = 16-12
2a = 4
a = 4/2
a =2 

Therefore,
a +d = 2+2 = 4
a+ 2d = 2+ 4 = 6
a+3d = 2+6 = 8
a +4d = 2 + 8= 10 
Answer is 4, 6, 8, 10

Answer 2

Answer:

2,8,10 ,4 in the answer in one