# Question #92422

Apr 1, 2017

I got $33$

#### Explanation:

$n$
$n + 1$
$n + 2$
$n + 3$
So we get:
$n + \left(n + 1\right) + \left(n + 2\right) + \left(n + 3\right) = 130$
rearrange and solve for $n$:
$4 n + 6 = 130$
$4 n = 124$
$n = \frac{124}{4} = 31$
So the third in the sequence will be $33$

Apr 1, 2017

$33$ (see explanation)

#### Explanation:

Since the sum consists of $4$ consecutive integers we can write the following and solve for $x$

$x + \left(x + 1\right) + \left(x + 2\right) + \left(x + 3\right) = 130$

$4 x + 6 = 130 \to 4 x + \cancel{6 - 6} = 130 - 6$

$4 x = 124 \to \cancel{\frac{4}{4}} x = \frac{124}{4}$

$x = 31$

Thus the four consecutive integers are as follows:

$31 + \left(31 + 1\right) + \left(31 + 2\right) + \left(31 + 3\right) = 130$

or

$31 , 32 , 33 , 34$

We can verify this by substituting $31$ into the equation we used to solve for $x$

$31 + 32 + 33 + 34 = 130$

$130 = 130$

Finally, since we were asked to find third number of the sequence, we know that our third number is $33$