Question

find three consecutive numbers whose sum is 48

Answer 1

We can demote the smallest even number by

n1=2n

So, the next consecutive even integers would be

n2=2(n+1)=2n+2, and
n3=2(n+2)=2n+4

So, the sum is:

n1+n2+n3=(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:

n1=14
n2=16
n3=18