Answer: 14
Explanation:
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
Hope this helps!
