the smallest 3 consecutive integers is added to twice the largest , the result so obtained is fifteen less than four times the middle integer. find the integers
Let the integers be x-1,x,x+1.
Then
(x-1)+2(x+1)=4x-15
Solve for x.
To solve this problem, let's represent the smallest integer as "n", the middle integer as "n+1", and the largest integer as "n+2".
According to the given information, the sum of the smallest 3 consecutive integers (n + n+1 + n+2) is equal to twice the largest integer (2 * (n+2)), and this sum is fifteen less than four times the middle integer (4 * (n+1) - 15).
Setting up the equation:
n + n+1 + n+2 = 2 * (n+2)
n + n+1 + n+2 = 4 * (n+1) - 15
Simplifying the equations:
3n + 3 = 2n + 4
3n + 3 = 4n + 4 - 15
Solving the equations:
3n + 3 = 2n + 4
3n - 2n = 4 - 3
n = 1
Now that we have the value of the smallest integer (n=1), we can find the other integers:
n+1 = 1+1 = 2
n+2 = 1+2 = 3
Therefore, the three consecutive integers are 1, 2, and 3.