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Four consecutive integers add up to 30. What is the greatest of these integers?

Sagot :

We can express the sum of these four consecutive integers in the following way:

[tex]n+(n+1)+(n+2)+(n+3)=30[/tex]

Then, we need to sum the like terms, that is n's with n's and the integers. Then, we have:

[tex](n+n+n+n)+(1+2+3)=30[/tex][tex]4n+6=30[/tex]

Then, solving for n, we can subtract 6 to both sides of the equation, and then divide by 4 to both sides of the equation too:

[tex]4n+6-6=30-6\Rightarrow4n=24\Rightarrow\frac{4}{4}n=\frac{24}{4}\Rightarrow n=6[/tex]

Since n = 6, the consecutive integers are:

[tex]6+(6+1)+(6+2)+(6+3)=6+7+8+9=30[/tex]

Therefore, the greatest of these integers is 9 (since we have 6, 7, 8, and 9).

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