The sum of three consecutive integers is minus, 51−51. Find the value of the middle of the three.

Let's denote the three consecutive integers as \( n-1 \), \( n \), and \( n+1 \), where \( n \) is the middle integer.

The sum of these three consecutive integers is given by:

\[
(n-1) + n + (n+1) = -51
\]

Now, simplify the left side:

\[
n - 1 + n + n + 1 = 3n
\]

So, the equation becomes:

\[
3n = -51
\]

To find the value of the middle integer \( n \), divide both sides by 3:

\[
n = \frac{-51}{3} = -17
\]

### Summary:
The value of the middle integer is \( \mathbf{-17} \).

Would you like further details or have any other questions?

Here are five related questions:
1. What are the other two integers in this sequence?
2. What would be the sum if the middle integer was -18 instead of -17?
3. If the sum of four consecutive integers was -51, what would be their values?
4. How would you find the sum of five consecutive integers where the middle one is given?
5. Can you find the sum of three consecutive integers if the middle one is 10?

**Tip:** When dealing with consecutive integers, remember that their sum is always divisible by the number of integers involved.

Discover how to find the middle integer of three consecutive integers that sum to -51. This problem involves basic arithmetic and linear equations, with step-by-step solution and explanation. Suitable for Grades 7-9 students learning about arithmetic sequences and equations.

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