Question 69102
If you assume that the first integer is N, then the next three integers are N+1, N+2, and N+3.

You need to know the sum of these four integers in a row.  The sum is N+(N+1)+ (N+2)+(N+3).  Since all the parentheses are preceded by a + sign, the parentheses can be removed without changing any of the terms in the parentheses.  The result is:

N+N+1+N+2+N+3 

and by adding like terms this sum is simplified to 4N+6

Next you need to find the sum of the remaining 8 numbers.  You can get this by adding all 12 numbers and then subtracting the sum of the four consecutive numbers we found above. Probably the quickest way to find the sum of all 12 numbers is to just add 21 through 32 on a calculator.  You can also find it by multiplying 12 times (21+32) and dividing that answer by 2.  [This method involves the rule that to add consecutive numbers in a series, add the first number in the series to the last number in the series, multiply that sum by the number of numbers in the series, and divide that result by 2.] In either case you should find that the sum of the 12 consecutive numbers in the series whose first number is 21 and whose last number is 32, is a total of 318.  If you subtract from 318 the sum of the four consecutive numbers, the remainder will be the sum of the remaining eight numbers.  Since the sum of the 4 consecutive numbers is 4N+6, the subtraction is:

318-(4N+6) = 318-4N-6 = 312-4N

This is the sum of the other eight numbers 

The problem tells you that the sum of the four consecutive numbers equals one-half the sum of the remaining eight numbers.  In equation form this is:

4N+6 = (1/2)*[312-4N) 

and this can initially be simplified by doing the multiplication on the right side to get:

4N+6 = 156-2N

First add 2N to both sides and then subtract 6 from both sides.  The result is:

6N = 150

Divide both sides by 6 and you find that N=25.  N is the first of the four consecutive numbers.  Therefore, the four consecutive numbers are 25, 26, 27, and 28.

Using a calculator you should find that the sum of 25+26+27+28 is 106.

The remaining eight numbers are 21, 22, 23, 24, 29, 30, 31, and 31.  Using a calculator again you should find that the sum of these eight numbers is 212.

And one-half of 212 is 106.  So the problem checks out.

Hope this helps you to understand how to work problems that involve consecutive integers.