Question 765882: The sum of 4 consecutive multiples of 12 have a sum of 216. What is the greatest of these numbers? Answer by suruman(21) (Show Source):
You can put this solution on YOUR website! There are 4 consecutive multiples of 12.
Let the consecutive numbers be:
12*(n-1), 12*(n),12*(n+1) and 12*(n+2).
Sum of the consecutive numbers equals 216.
Thus,
12 * {(n-1)+(n)+(n+1)+(n+2) } = 216
(3*n)+(n+2)=216/12 = 18
4*n + 2 = 18
4*n = 16
n=16/4 = 4
Hence the greatest of these numbers = 12*(n+2) = 12*(4+2) = 12*6 = 72.
