Question

The sum of three consecutive even numbers is 48. What is the smallest of these numbers?

Answer 1

The smallest number is `14`

Explanation:

Let:
x= the 1st con.even number
x+2=the 2nd con.even number
x+4=the 3rd con.even number

Add the terms and equate it with the total, 48
`x + (x+2)+(x+4)=48`, simplify
`x +x+2+x+4=48`, combine like terms
`3x+6=48`, isolate x
`x=(48-6)/3`, find the value of x
`x=14`

The 3 con.even numbers are the ff.:
`x=14` `->`the smallest number
`x+2=16`
`x+4=18`

Check:

`x +x+2+x+4=48`
`14+14+2+14+4=48`
`48=48`

Answer 2

`14`

Explanation:

We can demote the smallest even number by

`n_1 = 2n`

So, the next consecutive even integers would be

`n_2 = 2(n+1) = 2n + 2`, and
`n_3 = 2(n+2) = 2n +4`

So, the sum is:

`n_1+n_2+n_3 = (2n) + (2n+2) + (2n+4 )`

We are told that this sum is `48`, thus:

`(2n) + (2n+2) + (2n+4 ) = 48`

`:. 6n + 6 = 48`
`:. 6n = 42`
`:. n = 7`

And with `n=7`, we have:

`n_1 = 14`
`n_2 = 16`
`n_3 = 18`