# Four consecutive integers are such that the sum of the 2nd and 4th integers is 132. What are the four integers?

Sep 13, 2016

$64 , 65 , 66 , 67$

#### Explanation:

Suppose the integers are:

$n - 2 , n - 1 , n , n + 1$

Then we are given:

$132 = \left(n - 1\right) + \left(n + 1\right) = 2 n$

Dividing both ends by $2$ and transposing, we find:

$n = 66$

So the four integers are:

$64 , 65 , 66 , 67$

Sep 13, 2016

The four consecutive integers are 64,65, 66 and 67.

#### Explanation:

Consecutive integers are found by adding 1. For example, 2, 3 and 4 are consecutive integers.

For this problem:
Let the first =x

Let the second integer =x+1

Let the third integer =x+2

Let the fourth integer =x+3

The sum of the 2nd and 4th is
$x + 1 \textcolor{w h i t e}{a a a} + \textcolor{w h i t e}{a a a} x + 3$

$x + 1 + x + 3 = 132$

Combine like terms

$2 x + 4 = 132$

Subtract 4 from both sides.

$2 x + 4 - 4 = 132 - 4$
$2 x = 128$

Divide both sides by 2.
$\frac{2 x}{2} = \frac{128}{2}$
$x = 64$

The first integer is 64.
The 2nd is 65.
The 3rd is 66.
The 4th is 67.