# How do you find a set of four consecutive integers whose sum is equal to the sum of the next three consecutive integers immediately following them?

Jul 20, 2016

$9$, $10$, $11$, $12$

#### Explanation:

First let's identify the integers. The first one would be $n$, then $n + 1$, then $n + 2$, and so on. We are adding an additional number each time since the numbers are consecutive. Now let's write an equation.

$n + \left(n + 1\right) + \left(n + 2\right) + \left(n + 3\right) = \left(n + 4\right) + \left(n + 5\right) + \left(n + 6\right)$

Combine like terms.

$4 n + 6 = 3 n + 15$

Now solve for $n$. Subtract $3 n$.

$4 n - 3 n + 6 = \cancel{3 n - 3 n} + 15$

$n + 6 = 15$

Subtract $6$ from both sides.

$n \cancel{+ 6 - 6} = 15 - 6$

$n = 9$

We found that $n$ equals $9$. This means the first integer is $9$. We can plug this into $\left(n + 1\right)$, $\left(n + 2\right) , \mathmr{and}$(n+3)# to find the other three integers.

$\left(9 + 1\right) = 10$

$\left(9 + 2\right) = 11$

$\left(9 + 3\right) = 12$

The other three integers are $10$, $11$, and $12$.
Now we can check that these are the correct integers by finding the other three integers from the equation (the three integers that are the sum of the four known integers).

$\left(9 + 4\right) = 13$

$\left(9 + 5\right) = 14$

$\left(9 + 6\right) = 15$

Now that we know the other three integers, we can plug all known values into the original equation and see if the integers solved for are correct.

$n + \left(n + 1\right) + \left(n + 2\right) + \left(n + 3\right) = \left(n + 4\right) + \left(n + 5\right) + \left(n + 6\right)$

$9 + 10 + 11 + 12 = 13 + 14 + 15$

$42 = 42$

The sums match, so the integers are:

$9$, $10$, $11$, $12$