Question

There are 4 consecutive even integers that add up to –12. Which integers are they?

Answer 1

We have 4 even numbers which sum is equal to -12.

We can write the even numbers as 2n, 2(n+1) 2(n+2) and 2(n+3), where the index n is an integer.

If we add them, we get:

`\begin{gathered} 2n+2(n+1)+2(n+2)+3(n+3)=-12 \\ 2(n+n+1+n+2+n+3)=-12 \\ 2(4n+6)=-12 \\ 4n+6=-(12)/(2)=-6 \\ 4n=-6-6 \\ 4n=-12 \\ n=-(12)/(4)=-3 \end{gathered}`

As we have find the value of n, we can replace it in the definition of the integers, and find which ones satisfy the condition:

`\begin{gathered} n=-3\longrightarrow2n=2(-3)=-6 \\ 2(n+1)=2(-3+1)=2\cdot(-2)=-4 \\ 2(n+2)=2(-3+2)=2\cdot(-1)=-2 \\ 2(n+3)=2(-3+3)=2\cdot(-0)=0 \end{gathered}`

Answer: The numbers are: -6,-4,-2 and 0.