The sum of three consecutive integers is

−99. Find the three integers.

n-1 + n + n+1 = -99

3 n =-99
n = -33

-34 -33 -32

Let's assume the first integer is x.

The second consecutive integer would be x + 1.
And the third consecutive integer would be x + 2.

According to the given information, the sum of these three consecutive integers is -99.

So, we can write the equation as:
x + (x + 1) + (x + 2) = -99

Now, let's solve the equation to find the value of x.
Combining like terms, we have:
3x + 3 = -99

Subtracting 3 from both sides:
3x = -102

Dividing both sides by 3:
x = -34

So, the first integer is -34.
The second consecutive integer is -34 + 1 = -33.
The third consecutive integer is -34 + 2 = -32.

Therefore, the three integers are -34, -33, and -32.

To find the three consecutive integers, we can use the formula for the sum of an arithmetic sequence. The formula states that the sum of an arithmetic sequence can be found by taking the average of the first and last terms, and multiplying it by the number of terms.

Let's represent the first consecutive integer as x. The next two consecutive integers would be (x + 1) and (x + 2).

According to the problem, the sum of these three consecutive integers is -99. Therefore, we can write the equation:

x + (x + 1) + (x + 2) = -99

Now, we can solve this equation for x to find the value of the first integer:

3x + 3 = -99

Subtract 3 from both sides:

3x = -102

Divide by 3:

x = -34

So, the first integer is -34. The next two consecutive integers would be -34 + 1 = -33 and -34 + 2 = -32.

Therefore, the three consecutive integers are -34, -33, and -32.