Question

Question #92422

Answer 1

I got `33`

Explanation:

Call your numbers:
`n`
`n+1`
`n+2`
`n+3`
So we get:
`n+(n+1)+(n+2)+(n+3)=130`
rearrange and solve for `n`:
`4n+6=130`
`4n=124`
`n=124/4=31`
So the third in the sequence will be `33`

Answer 2

`33` (see explanation)

Explanation:

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

`x+(x+1)+(x+2)+(x+3)= 130`

`4x+6=130 -> 4x+cancel(6-6)=130-6`

`4x=124 -> cancel(4/4)x=124/4`

`x=31`

Thus the four consecutive integers are as follows:

`31+(31+1)+(31+2)+(31+3)=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`