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Inverted Sum

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by linkinpark » Wed Dec 23, 2009 6:53 am
maihuna wrote:S IS THE SUM OF 1/N, WHERE 101<=N<=150,WHAT'S S?
is it 1/6275?
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by maihuna » Wed Dec 23, 2009 8:09 am
linkinpark wrote:
maihuna wrote:S IS THE SUM OF 1/N, WHERE 101<=N<=150,WHAT'S S?
is it 1/6275?
How?
Charged up again to beat the beast :)
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by linkinpark » Wed Dec 23, 2009 8:17 am
Now I'm slightly confused whether I've understood the question but for 1/6275, I sumed up 101 to 150 and inverted it.

I think it should be like 1/101 + 1/102 + 1/103 + .... + 1/150 which approximately should be 0.01.

though it's good question
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by Stuart@KaplanGMAT » Wed Dec 23, 2009 6:44 pm
linkinpark wrote:Now I'm slightly confused whether I've understood the question but for 1/6275, I sumed up 101 to 150 and inverted it.

I think it should be like 1/101 + 1/102 + 1/103 + .... + 1/150 which approximately should be 0.01.

though it's good question
You're right.. the question was likely meant to be the first answer that you posted, although it actually means the second one that you posted, which is pretty much impossible to answer without a computer or a LOT of time and scrap paper.

When you post a question on this board, please always do 3 things:

1) make sure you post the entire question, correctly;
2) include the answer choices; and
3) cite the source.

Assuming that the question is supposed to be:

What's the inverse of 1/(101+102+103+...+150), we can solve fairly quickly:

There are 50 terms from 101 to 150.

To calculate the sum of a set of consecutive numbers, we multiply the number of terms by the average of the set.

The average of a set of consecutive numbers is (smallest + biggest)/2.

So, our sum is:

50 * (101 + 150)/2 = (50 * 251)/2 = 12550/2 = 6275

and the final answer is:

1/6275.
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