Question

How do you find three consecutive odd integers such that twice the sum of the first and third integers is 21 more than the second integer?

Answer 1

The three consecutive odd numbers are `5, 7, 9`

Explanation:

Let
`color(white)("XXX")`first odd number `= color(red)(2n-1)`
`color(white)("XXX")`second odd number `=color(blue)(2n+1)`
`color(white)("XXX")`third odd number `=color(brown)(2n+3)`

We are told
`color(white)("XXX")2( color(red)((2n-1)) + color(brown)((2n+3)) ) = color(blue)((2n+1)) + 21`

Simplifying
`color(white)("XXX")8n+4 = 2n+22`

`color(white)("XXX")6n = 18`

`color(white)("XXX")n=3`

So the three odd numbers are
`color(white)("XXX")`first `= 2n-1 = 2(3)-1 = 5`
`color(white)("XXX")`second `= 2(3)+1 = 7`
`color(white)("XXX")`third `= 2(3)+3 = 9`