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# M is the sum of the reciprocals of the consecutive integers

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

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22 Jun 2015, 18:20
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Re: M is the sum of the reciprocals of the consecutive integers [#permalink]

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24 Jun 2015, 20:02
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Absolute Value/Modulus tag is not required in this
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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Explanation needed - Reciprocals [#permalink]

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16 Jul 2015, 17:07
Please could you help me with this one:

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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Explanation needed - Reciprocals [#permalink]

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16 Jul 2015, 17:09
cuckio wrote:
Please could you help me with this one:

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

Please search for your question before you post. Follow the rules for posting (links in my signatures) and choose the correct forum to post your question.

The same question has been discussed at : m-is-the-sum-of-the-reciprocals-of-the-consecutive-integers-143703.html

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

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01 Jun 2016, 07:39
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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M is the sum of the reciprocals of the consecutive integers [#permalink]

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01 Aug 2016, 17:11
1/200+1/300=1/120
(1/120)/2=1/240
(1/240)100=5/12=~M
1/3<5/12<1/2
only answer A works

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M is the sum of the reciprocals ... [#permalink]

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04 Sep 2016, 08:56
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
(0 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 ... [#permalink]

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04 Sep 2016, 10:16
Manonamission 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
(0 1/7<M<1/5
(D) 1/9<M<1/7
(E) 1/12<M<1/9

Minimum value of M > 1/300 +1/300 + .. 100 times = 100/300 =1/3

Max Value of M < 1/200 +1/200 + .. 100 times = 100/200 = 1/2.

Hence, Answer is A.
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Re: M is the sum of the reciprocals ... [#permalink]

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04 Sep 2016, 11:30
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
(0 1/7<M<1/5
(D) 1/9<M<1/7
(E) 1/12<M<1/9

1/200+1/300=5/600=1/120
(1/120)/2=1/240
100*1/240=5/12=M
1/3<5/12/<1/2
A

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Re: M is the sum of the reciprocals ... [#permalink]

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04 Sep 2016, 13:12
abhimahna wrote:
Manonamission 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
(0 1/7<M<1/5
(D) 1/9<M<1/7
(E) 1/12<M<1/9

Minimum value of M > 1/300 +1/300 + .. 100 times = 100/300 =1/3

Max Value of M < 1/200 +1/200 + .. 100 times = 100/200 = 1/2.

Hence, Answer is A.

I would like to see another explanation.

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Re: M is the sum of the reciprocals ... [#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 [#permalink]

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04 Sep 2016, 22:31
Manonamission 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
(0 1/7<M<1/5
(D) 1/9<M<1/7
(E) 1/12<M<1/9

Merging topics. Please search before posting.
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M is the sum of the reciprocals of the consecutive integers [#permalink]

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24 Oct 2016, 12:11
I wanted to take this approach but I'm not sure how correct the formula is...

$$Sn = (100/2) * [(2)(1/300) + (100-1)(1/n)]$$

The sum isn't coming out to a value between 1/3 and 1/2, but I normally use this formula when it comes to summation for sequences. I think my issue is figuring out what the relationship (or increment) should be for the n, n + 1, etc.

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

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11 Aug 2017, 11:41
sum of consecutive integers = mean * number of n integers in the sequence.

mean = (first + last) / 2 = (201 + 300) /2 ~ 1/ 250
1/ 250 * 100 (number of integers in the sequence) = 100/250 = 2/ 5

Only A matches

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Re: M is the sum of the reciprocals of the consecutive integers [#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   [#permalink] 14 Aug 2017, 13:14

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