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M is the sum of the reciprocals of the consecutive integers from 201

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Re: M is the sum of the reciprocals of the consecutive integers from 201  [#permalink]

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New post 04 Sep 2016, 20:55
Hi Manonamission,

In this prompt, the answer choices are "ranges"; this usually means that there's a way to avoid doing lots of math and instead use patterns and logic to save you time. You can actually stop working once you figure out the minimum value...

Since 1/300 < 1/201 and the sum of those 100 terms would be (100)(1/300) = 1/3 AT THE MINIMUM, the only answer that's possible would be A. The extra work that you could do to find the maximum value just confirms what the sum would be at that 'level', but but that work is unnecessary.

As you continue to study, be mindful of how the answer choices are written - they can sometimes provide a huge hint into the fastest way to answer the question.

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Re: M is the sum of the reciprocals of the consecutive integers from 201  [#permalink]

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New post 14 Aug 2017, 13:14
Bunuel wrote:
M is the sum of the reciprocals of the consecutive integers from 201 to 300, inclusive. Which of the following is true?

(A) 1/3 < M < 1/2
(B) 1/5 < M < 1/3
(C) 1/7 < M < 1/5
(D) 1/9 < M < 1/7
(E) 1/12 < M < 1/9

Given that \(M=\frac{1}{201}+\frac{1}{202}+\frac{1}{203}+...+\frac{1}{300}\). Notice that 1/201 is the larges term and 1/300 is the smallest term.

If all 100 terms were equal to 1/300, then the sum would be 100/300=1/3, but since actual sum is more than that, then we have that M>1/3.

If all 100 terms were equal to 1/200, then the sum would be 100/200=1/2, but since actual sum is less than that, then we have that M<1/2.

Therefore, 1/3<M<1/2.

Answer: A.


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Re: M is the sum of the reciprocals of the consecutive integers from 201  [#permalink]

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New post 14 Oct 2017, 16:44
This is how I solved this one:

1/200=0.005
1/300=0.003approx

Average of both is 0.004. There are 100 numbers from 201 to 300 hence the sum of their reciprocals must be very close to 0.004*100=0.4

0.4 is between 1/3 and 1/2.

Answer=A

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Re: M is the sum of the reciprocals of the consecutive integers from 201  [#permalink]

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New post 05 Nov 2017, 12:21
Roughly the median of the set would be 1/250 and there are around 100 nos. in the set. SO the sum would be roughly 100*(1/250)

So option A
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M is the sum of the reciprocals of the consecutive integers from 201  [#permalink]

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New post 17 Jan 2018, 19:07
Bunuel chetan2u niks18
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M is the sum of the reciprocals of the consecutive integers from 201 to 300, inclusive. Which of the following is true?

If all 100 terms were equal to 1/200, then the sum would be 100/200=1/2, but since actual sum is less than that, then we have that M<1/2.


Can I calculate total sum using standard formula for AP:

The general sum of a n term AP with common difference d is given by \(\frac{n}{2}(2a+(n-1)d)\)

Here: difference is zero and first term is 1/300. Total no of terms =100

Or, did you simply factor out 1/300 common and multiplied it by no of terms ie 100?
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Re: M is the sum of the reciprocals of the consecutive integers from 201  [#permalink]

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New post 18 Jan 2018, 18:42
chetan2u

I apologize that I did not put forth my query well.

My query was: What is the best way to simplify \(\frac{1}{200}\) + \(\frac{1}{200}\)+ .. Total 100 times

Does my later approach of simplification in the earlier post holds good?
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Re: M is the sum of the reciprocals of the consecutive integers from 201  [#permalink]

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New post 18 Jan 2018, 19:16
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adkikani wrote:
chetan2u

I apologize that I did not put forth my query well.

My query was: What is the best way to simplify \(\frac{1}{200}\) + \(\frac{1}{200}\)+ .. Total 100 times

Does my later approach of simplification in the earlier post holds good?



Yes ...
In the formula when you put d as 0, it boils down to n/2*2a=na..
n is 100 and a is 1/200, same as taking out 1/200 outside
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Re: M is the sum of the reciprocals of the consecutive integers from 201  [#permalink]

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New post 04 May 2018, 04:48
Hi Bunuel! Would be gratified if you can look at my solution below and inform if its correcrt or not.

Avg of #= S/n
Avg= [(1/250)+(1/251)]/2
Avg= 501/(250*251*2)
Sum= Avg*n
S= 100* [501/(250*251*2)]
S= 501/1255
Assuming 501=500
S=100/251
S=1/251*10^(-2)
S=1/2.51
Hence option A
Re: M is the sum of the reciprocals of the consecutive integers from 201 &nbs [#permalink] 04 May 2018, 04:48

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