Answer:
89406, 89408 and 89410.
Step-by-step explanation:
Let n be the any even integer.
So n+2 will be 2nd even integer and n+4 will be 3rd even integer.
We are told that twice the first integer added to the third is 268,222.
Twice the first integer will be 2*n.
We can represent this information in an equation as:
[tex]2n+(n+4)=268222[/tex]
[tex]2n+n+4=268222[/tex]
[tex]3n+4=268222[/tex]
Let us subtract 4 from both sides of our equation.
[tex]3n+4-4=268222-4[/tex]
[tex]3n=268218[/tex]
Let us divide both sides of our equation by 3.
[tex]\frac{3n}{3}=\frac{268218}{3}[/tex]
[tex]n=89406[/tex]
Therefore, our 1st even integer will be 89406.
2nd even integer = n+2
[tex]n+2=89406+2=89408[/tex]
3rd even integer = n+4
[tex]n+4=89406+4=89410[/tex]
Therefore, our three consecutive even integers will be 89406, 89408 and 89410.
