I found this easy problem in the grade 8th entrance exam... try it :-) Let \(\large x_1, x_2, x_3 \cdots x_n\) be a sequence such that \(\large \sum \limits_{i = 1}^{n} (x_i - 3) = 170 \) and \(\large \sum \limits_{i = 1}^{n} (x_i - 6) = 50\). What is the value of \(n\)?

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I found this easy problem in the grade 8th entrance exam... try it :-) Let \(\large x_1, x_2, x_3 \cdots x_n\) be a sequence such that \(\large \sum \limits_{i = 1}^{n} (x_i - 3) = 170 \) and \(\large \sum \limits_{i = 1}^{n} (x_i - 6) = 50\). What is the value of \(n\)?

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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0k ?
Okay, it's not a hard problem, but it is tedious and requires knowledge about summations that is barely touched in many algebra 2 classes.
Multiply the first equation by \(-2\) and then add it to the second equation.

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$$\sum \limits_{i=1}^nx_i=p\\ \sum\limits_{i=1}^nx_i-\sum\limits_{i=1}^n3=170 \implies p-3n=170\qquad (1)\\ \sum\limits_{i=1}^nx_i-\sum\limits_{i=1}^n6=50\implies p-6n=50 \qquad(2)$$ (1)-(2) 3n=120 => n=40
@BAdhi :-D

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