Question

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

Answer 1

`64, 65, 66, 67`

Explanation:

Suppose the integers are:

`n-2, n-1, n, n+1`

Then we are given:

`132 = (n-1) + (n+1) = 2n`

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

`n = 66`

So the four integers are:

`64, 65, 66, 67`

Answer 2

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+1color(white)(aaa)+color(white)(aaa)x+3`

`x+1+x+3=132`

Combine like terms

`2x+4=132`

Subtract 4 from both sides.

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

Divide both sides by 2.
`(2x)/2=128/2`
`x=64`

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