Answer:
6, 8, 10.
Step-by-step explanation:
Let n be any integer.
Then our first even integer will be [tex]2n[/tex]
And the consecutive integers will be [tex]2n+2[/tex] and [tex]2n+4[/tex]
We want to find the integers such that the sum of twice the smallest number (2n) and three times the largest number (2n+4) is 42. In other words:
[tex]2(2n)+3(2n+4)=42[/tex]
Solve for n. Distribute:
[tex]4n+6n+12=42[/tex]
Combine like terms:
[tex]10n+12=42[/tex]
Subtract 12 from both sides:
[tex]10n=30[/tex]
Divide both sides by 10:
[tex]n=3[/tex]
So, the value of n is 3.
This means that our first even integer is 2(3) or 6.
So, our sequence of integers is: 6, 8, 10.
And we're done!
Notes:
We use 2n because anything integer multiplied by 2 ensures that the resulting number is even. Starting out, n can be either even or odd, but by multiplying it by 2, we will get an even number.
