Question

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

Answer 1

18 , 20 , 22 and 24.

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.