Find 4 consecutive even integers such that the sum of the first and three times the fourth is equal to 48 more than the sum of 2nd and 3rd integers

18, 20, 22 and 24.

Step-by-step explanation:

Let the numbers be n, n + 2, n + 4 and n + 6.

So we have the following equation:

n + 3 (n + 6) = 48 + n + 2 + n + 4

n + 3n + 18 = 48 + 2n + 6

4n + 18 = 2n + 54

2n = 36

n = 18

So the numbers are 18, 20, 22 and 24.