Find three consecutive even integers such that the sum of the smallest integer and twice the median is 20 more than the largest integer

Let x, x+2 & x+4 be the 3 integers.

x + 2 (x+2) = (x+4) + 20

x + 2x + 4 = x + 4 + 20

3x + 4 = x + 24

3x - x = 24 - 4

2x = 20

x = 20/1

x=10 ans. for the smallest integer.

10 + 2 = 12 for the middle integer.

10 + 4 = 14 ans. for the largest integer.

To check:

10 + 2*12 = 14 + 20

10 + 24 = 34

34 = 34