Find three consecutive odd integers such that four times the middle integer is two more than the sum of the first and third.
n, n+2, n+4
4(n+2) = n + n+4 + 2
To find three consecutive odd integers, let's start by assigning variables to them. Let's say the first odd integer is n, the second (middle) odd integer is n + 2, and the third odd integer is n + 4.
According to the problem, four times the middle integer is two more than the sum of the first and third. We can write this as an equation:
4(n + 2) = (n + n + 4) + 2
Now, let's simplify the equation:
4n + 8 = 2n + 6 + 2
4n + 8 = 2n + 8
Next, let's move all the variables to one side:
4n - 2n = 8 - 8
2n = 0
Finally, we can solve for n:
2n = 0
n = 0 / 2
n = 0
Now that we have the value of n, we can find the three consecutive odd integers:
First odd integer: n = 0
Second (middle) odd integer: n + 2 = 0 + 2 = 2
Third odd integer: n + 4 = 0 + 4 = 4
Therefore, the three consecutive odd integers are 0, 2, and 4.