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Manager  Joined: 02 Dec 2012
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M is the sum of the reciprocals of the consecutive integers from 201  [#permalink]

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

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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

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.

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

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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

##### General Discussion
Intern  Joined: 21 Oct 2012
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GMAT Date: 01-19-2013
Re: M is the sum of the reciprocals of the consecutive integers from 201  [#permalink]

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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.

WOW, that is elegant!
Bunuel please suggest if we can use the next assumption:
Arithmetic mean of the elements a1 and a100= 501/(300*2*201). Sum of all elements = 100*Arithmetic mean=167/(6*67)= 167/402 , which is definitely more than 1/3. A
Thanks
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Re: M is the sum of the reciprocals of the consecutive integers from 201  [#permalink]

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Whoah! I would never approach that question this way.
Thank you for this post!
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Re: M is the sum of the reciprocals of the consecutive integers from 201  [#permalink]

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mbaiseasy wrote:
Whoah! I would never approach that question this way.
Thank you for this post!

Now that you know you would .
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Re: M is the sum of the reciprocals of the consecutive integers from 201  [#permalink]

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AbhiJ wrote:
mbaiseasy wrote:
Whoah! I would never approach that question this way.
Thank you for this post!

Now that you know you would .

Similar question to practice: if-k-is-the-sum-of-reciprocals-of-the-consecutive-integers-145365.html

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

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

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I did this similar to the post above.

Between 201 and 300 there are 100 numbers (300-201+1 = 100). Since the integers 201 to 300 are consecutive, the middle number is ((201+300)/2) which is roughly 250. Therefore, the sum is approximately:

(1/250)*100 = (1/25) * (1/10) * 100 = (1/25) * 10 = (0.04)*10 = 0.40

And by inspecting the answers 1/3 < 0.4 < 1/2 so I get (A).
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Re: M is the sum of the reciprocals of the consecutive integers from 201  [#permalink]

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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

There is also an another easier way to solve this problem - using sum and average concept.
1. The numbers are reciprocal of consecutive numbers.
Total number of items = 100 (201-300 +1 ) => Middle term is 250 and the number is the reciprocal which is 1/250.

2. Now we know for consecutive numbers and odd number of terms, Average = Middle number (also Average = Median) and Average in general is Sum / Number of terms. In this case average = 1/250.

Hence, we have Average (1/250) = Sum / 100.
Solving this sum = 100/250 = 0.4

From the answer choices, 0.4 is between 1/3 and 1/5.
Math Expert V
Joined: 02 Sep 2009
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Re: M is the sum of the reciprocals of the consecutive integers from 201  [#permalink]

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coolparthi 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

There is also an another easier way to solve this problem - using sum and average concept.
1. The numbers are reciprocal of consecutive numbers.
Total number of items = 100 (201-300 +1 ) => Middle term is 250 and the number is the reciprocal which is 1/250.

2. Now we know for consecutive numbers and odd number of terms, Average = Middle number (also Average = Median) and Average in general is Sum / Number of terms. In this case average = 1/250.

Hence, we have Average (1/250) = Sum / 100.
Solving this sum = 100/250 = 0.4

From the answer choices, 0.4 is between 1/3 and 1/5.

First of all the average of consecutive integers from 201 to 300, inclusive is (201+300)/2=250.5 not 250.

Next, set {1/201, 1/202, 1/203, ..., 1/300} is NOT evenly spaced, thus your method is an approximation (the formula you apply is for an evenly spaced set).

Hope it's clear.
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Re: M is the sum of the reciprocals of the consecutive integers from 201  [#permalink]

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5
There is another simple way to do this problem, when reciprocals of numbers from 201 to 300 are added...then it means the denominators are in Arithmetic progression...so 1/201, 1/202, ---, 1/300 are in Harmonic progression. Sum of all these numbers is (# of terms)*(2*First term*Last term)/(First term+Last term)
which is (2*(1/201)*(1/300))/(1/201+1/300)...simplifying gives us 200/501 ~ 200/500 = 1/2.5 so lies between 1/3 and 1/2 Ans A.
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Re: M is the sum of the reciprocals of the consecutive integers from 201  [#permalink]

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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.

Wow, that was an awesome explanation. Very helpful.
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Joined: 09 Feb 2013
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Re: M is the sum of the reciprocals of the consecutive integers from 201  [#permalink]

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Will the answer pattern remain the same for the varies interval nos. ?

Suppose , if M is the sum of reciprocals of the cons. integers from 301 to 400 then the answer will be 1/4[m]1/3 ?
Is the above generalisation correct ?
Intern  Joined: 15 Apr 2014
Posts: 1
Re: M is the sum of the reciprocals of the consecutive integers from 201  [#permalink]

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kshitij89 wrote:
Will the answer pattern remain the same for the varies interval nos. ?

Suppose , if M is the sum of reciprocals of the cons. integers from 301 to 400 then the answer will be 1/4[m]1/3 ?
Is the above generalisation correct ?

your generalization would be correct but as you notice this form of pattern matching is a quick way to get to the right answer when you are short of time AND yes this holds true for 1/101 to 200, 1/201 to 300, 1/301 to 400 .......etc.
Hope this helps.

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

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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

Solution:

Let's first analyze the question. We are trying to find a potential range for M, and M is equal to the sum of the reciprocals from 201 to 300, inclusive. Thus, M is:

1/201 + 1/202 + 1/203 + …+ 1/300

There is no way the GMAT would ever expect us to do this math, and that is exactly why the answer choices in are in the form of an inequality. Thus, we do not need to know the EXACT value of M. The easiest way to determine the RANGE of M is to use easy numbers that can quickly be manipulated.

Note that 1/200 is greater than each of the addends and that 1/300 is less than or equal each of the addends. Therefore, instead of trying to add together 1/201 + 1/202 + 1/203 + …+ 1/300, we are instead going to add 1/200 one hundred times and 1/300 one hundred times. These two sums will give us a high estimate of M and a low estimate of M. Again, we are adding 1/200, one hundred times, and 1/300, one hundred times, because there are 100 numbers from 1/201 to 1/300.

Instead of actually adding each one of these values one hundred times, we will simply multiply each value by 100. We have:

1/300 x 100 = 1/3

1/200 x 100 = 1/2

We see that M is between 1/3 and 1/2.

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GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: M is the sum of the reciprocals of the consecutive integers from 201  [#permalink]

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

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Here is a detailed explanation in video for this question

_________________ Re: M is the sum of the reciprocals of the consecutive integers from 201   [#permalink] 19 Dec 2018, 02:16

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