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$$\sum_{i=x}^{2x} 5 - \sum_{k=2x+1}^{5x+2} 5 = -25$$

How can I discover the value of X?

Thanks in advance!

EDIT: Turns out it was 5x+2, not 5x+1. My bad, apologies for the trouble.

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Dividing by 5 gives (x+1)-(3x+1) = -5, so x=2.5 which is not an integer, so you have a mistake somewhere. – Derek Jennings Oct 31 '10 at 20:23
I'm sorry, I corrected the problem. – Qosmo Oct 31 '10 at 20:43

2 Answers 2

up vote 1 down vote accepted

You asked a similar question, so I think you didn't understand Derek's comment. Here we need to count how many integers there are between $x$ and $2x$ and between $2x+1$ and $5x+2$. Since you are not familiar with these things, try to write all possible numbers. You'll get something like $$x,\ x+1, \ldots, \ x+ (x-1), \ 2x$$ and $$2x+1, \ldots, \ 2x + (x-1), \ 3x, \ 3x+1, \ldots, \ 3x+ (x-1), \ 4x, $$ $$4x+1, \ldots, \ 4x +(x-1), \ 5x, \ 5x+1, \ 5x+2.$$ Now it's easy to see that you have $x+1$ numbers between $x$ and $2x$ and $3x+2$ between $2x+1$ and $5x+2$.

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Use the fact that

$\sum_{i=x}^{2x} 5 = 5 \cdot (2x - x) = 5x$


$\sum_{k=2x+1}^{5x+1} 5 = 5 \cdot (5x+1-2x-1) = 15x $

Then, the equation becomes

$5x - 15x = -25$

which is easy to solve, you just get $x = 25/10$ or in simpler terms $x = 5/2$

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I'm sorry, I corrected the problem. And this is what I was looking for, I couldn't remember what to do in this situation. Much appreciated. – Qosmo Oct 31 '10 at 20:42
You might want to update your answer just for the sake of correctness. Thanks again. – Qosmo Oct 31 '10 at 20:51
Hey downvoters: if an answer you see is "wrong", you might want to check first if the OP asked the "right" question before downvoting. Don't be jerks. – J. M. Oct 31 '10 at 23:21
@J.M. : I downvoted this answer, since $\sum_{i=x}^{2x}5= 5(x+1)$ not $5x$. After that the OP created a similar question making the same mistake. – Nuno Nov 1 '10 at 0:17
@Nuno: In any event, it's always a good idea to leave a comment on why you downvoted something. – J. M. Nov 1 '10 at 0:33

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