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 $n$?

Question

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 $n$?

goformit100 · Answer

0k ?

anonymous · Answer

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.

anonymous · Answer

Multiply the first equation by $-2$ and then add it to the second equation.

badhi · Answer

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

parthkohli · Answer

@BAdhi :-D