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During a test I was given the following question:

What is $$10+15+20+25+...+1490+1495+1500=?$$

After process of elimination and guessing, $ came up with an answer. Can anyone tell me a simple way to calculate this problem without having to try to actually add the numbers?

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    http://en.wikipedia.org/wiki/Arithmetic_progression2010-12-21
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    Please change to title to something more descriptive.2010-12-21
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    I guess Ronnie chose to ignore... edited the title and tag.2010-12-22
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    oh im sorry i didn't see your comment2010-12-22
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    No worries. Please just keep in mind to give more descriptive titles to the questions you ask, in the future.2010-12-22

4 Answers 4

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Hint 1 - Is there anything you can factor out of all the numbers?

Hint 2 - Do you know how to sum $1 + 2 + \cdots + n$ in a simple way?

EDIT - Your process in the comments is not wrong, but I don't feel like it's very intuitive.

I do think it's immediately clear that all the numbers in your sum above are divisible by 5. That is, we can rewrite $$10 + 15 + 20 + \cdots + 1500 = 5(2 + 3 + 4 + \cdots + 300)$$

Maybe not as obvious, but incredibly useful to know, is that $$1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}$$

Finally notice $2 + 3 + \cdots + 300$ is almost $1 + 2 + \cdots + 300$, and you can use the above formula to finish the problem.

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    would it be 225745?2010-12-21
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    That looks right to me. You should post your process, too, if you're interested in feedback for that.2010-12-21
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    ok well what i did was (after doing some reasearch) i used this formula n(a1+an)/2 substituded the values (299*(10+1500)/2) and got 451490/2 which gave me 2257452010-12-21
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    This isn't really what my answer was intending. I'll edit with the process I was thinking.2010-12-22
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    ok thank you for helping me with this2010-12-22
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    hello thank you for your process. I used your formula and for n I used 300. so 300*301=90300. 90300/2=45150. I then multiplied that by 5 but i got 225750. What am i doing wrong?2010-12-22
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    @Ronnie: Think about how the sum from 1 to 300 differs from the sum from 2 to 300. (This is what Hans was implicitly asking you to think about when he said "almost".)2010-12-22
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HINT: $(10 + 1500) + (15 + 1495) + (20 + 1490) + \cdots + (750 + 760) + 755$.

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This is probably the method C. F. Gauss would have used ;)


Factor out 5:

$5(2+3+4+5+...+297+298+299+300)$

Notice: $302/2 = 151$

$5((2+300)+(3+299)+(4+298)+(5+297)+...(149+153)+(150+152)+(151))$

$5(302+302+302+302+...+302+302+151)$

There are $(150-2+1)$ 302s, so:

$5((150-2+1)(302) + 151)$

$5((149)(302) + 151)$

$5(44998 + 151)$

$5(45149)$


The answer is $225745$.

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    lol thanks and i did a project on C.F. Gauss haha2010-12-22
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    @Ronnie Oh really? What's his last name?2010-12-23
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Hint 3 (Gauss): $10 + 1500 = 15 + 1495 = 20 + 1490 = \cdots = 1500 + 10$.