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Algebra Difficulty 1.1 Junior Find the answer Canada

If the average of $n-3$ , $n-1$ , $n+1$ , and $n+3$ is 17, the value of $n$ is

Pick one

Solution

The average of the $4$
expressions is $17$ , and so their sum
is $4\times17=68$ .

Thus, $(n-3)+(n-1)+(n+1)+(n+3)=68$ or
$4n=68$ , and so $n=\dfrac{68}{4}=17$ .

Can you see why the value of $n$ is
equal to the average?

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Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.