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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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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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10 Jul 2017, 10:57
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.

Simply awesome solution....how do i reach where u r...Q60
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Re: M is the sum of the reciprocals of the consecutive integers from 201  [#permalink]

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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.

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

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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.

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

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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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17 Jan 2018, 19:07
Bunuel chetan2u niks18
Quote:
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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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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18 Jan 2018, 19:16
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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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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
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M is the sum of the reciprocals of the consecutive integers from 201  [#permalink]

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13 Aug 2018, 05:35
mike34170 wrote:
I'd handle this differently:

Numbers range from 201 to 300, so the average is approximatively 250.

Then, the sum of all reciprocals equals to 100 times 1/250, i.e. 100/250 ---> 1/2,5

This is an approach that has proved well for me, thanks a lot for your input.
It works as well on another question proposed by bunuel : https://gmatclub.com/forum/if-k-is-the-sum-of-reciprocals-of-the-consecutive-integers-145365.html
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Re: M is the sum of the reciprocals of the consecutive integers from 201  [#permalink]

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30 Oct 2018, 22:04
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.

I have another approach .

Reciprocal of consecutive integers (a-----b) forms a harmonic progression
a < HMean < b
HMean = M / 100

a= 1/300 , b=1/200

1/300*100 < M < 1/200*100

Is it right approach ?>
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M is the sum of the reciprocals of the consecutive integers from 201  [#permalink]

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26 Nov 2018, 06:32
mike34170 wrote:
I'd handle this differently:

Numbers range from 201 to 300, so the average is approximatively 250.

Then, the sum of all reciprocals equals to 100 times 1/250, i.e. 100/250 ---> 1/2,5

Although you do get the answer, it is just a coincidence as the correct way to solve is given above by bunel.
Let me explain why this doesn't work using a simpler example!

Let's say we have the following sum:

1/1 + 1/2 + 1/3 + 1/4 + 1/5.

If we solve this with a calculator, we get 2.283

If we use the approach you suggested for this one, we would have the middle value (1/3) multiplied by 5.

(1/3) * 5 = 5/3 = 1.66

This is not the same as 2.283. So the method doesn't really work when we have fractions.

If we have a sum that doesn't involve fractions, the method does work. For example:

1+2+3+4+5 = 15

3 * 5 = 15

However, this doesn't work with fractions.
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Re: M is the sum of the reciprocals of the consecutive integers from 201  [#permalink]

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19 Dec 2018, 02:16
Here is a detailed explanation in video for this question

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

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25 Mar 2019, 09:17
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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?

(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

We want to find 1/201 + 1/202 + 1/203 + . . . + 1/299 + 1/300

NOTE: there are 100 fractions in this sum.

Let's examine the extreme values (1/201 and 1/300)

First consider a case where all of the values are equal to the smallest fraction (1/300)
We get: 1/300 + 1/300 + 1/300 + ... + 1/300 = 100/300 = 1/3
So, the original sum must be greater than 1/3

Now consider a case where all of the values are equal to the biggest fraction (1/201)
In fact, let's go a little bigger and use 1/200
We get: 1/200 + 1/200 + 1/200 + ... + 1/200 = 100/200 = 1/2
So, the original sum must be less than 1/2

Combine both cases to get 1/3 < M < 1/2

Cheers,
Brent
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Re: M is the sum of the reciprocals of the consecutive integers from 201   [#permalink] 25 Mar 2019, 09:17

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