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by Brent@GMATPrepNow » Tue Apr 12, 2016 6:57 am
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 = A

Cheers,
Brent
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by [email protected] » Tue Apr 12, 2016 10:22 am
Hi ezzeldin.khaled79,

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.

Brent's already pointed out the easiest way to figure out the minimum and maximum values of the sum of the reciprocals. You can actually stop working once you figure out the minimum though:

Since 1/300 < 1/201 and the sum of those 100 terms would be 1/3 AT THE MINIMUM, the only answer that's possible would be A. The extra work that Mitch did just confirms the maximum value of the sum, but it's 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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by OptimusPrep » Thu Apr 14, 2016 7:35 pm
[email protected] wrote:Can you guys explain the right answer for this Question?

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M is the sum of reciprocals of consecutive integers from 201 to 300. We need to find the range of M.
Number of terms in M = (300 - 201) + 1 = 100

M = 1/201 + 1/202 + ... + 1/300
If we need to find the greatest value of M, we need to maximize all the terms.
We can maximize each term by reducing the denominator

Hence M < 1/201 + 1/201 + 1/201 + ... 1/201
M < 100/201 - (i)


If we need to find the smallest value of M, we need to minimize all the terms.
We can minimize each term by increasing the denominator

Hence M > 1/300 + 1/300 + ... + 1/300
M > 100/300 Or M > 1/3 - (ii)

From (i) and (ii), we can say
1/3 < M < 100/201
We know that 100/201 < 100/200
Hence M < 100/200 or M < 1/2

Therefore 1/3 < M < 1/2

Option A
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by Matt@VeritasPrep » Fri Apr 15, 2016 1:27 pm
We know that

1/200 + 1/200 + ... + 1/200 > 1/201 + 1/202 + ... + 1/300 > 1/300 + 1/300 + ... + 1/300

or

100*(1/200) > 1/201 + 1/202 + ... + 1/300 > 100*(1/300)

or

1/2 > 1/201 + 1/202 + ... + 1/300 > 1/3

and we're set!
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