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M is the sum of the reciprocals of the consecutive integers

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M is the sum of the reciprocals of the consecutive integers [#permalink] New post 06 Dec 2012, 09:42
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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 [#permalink] New post 06 Dec 2012, 09:45
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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.

Answer: A.
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Re: M is the sum of the reciprocals of the consecutive integers [#permalink] New post 11 Dec 2012, 10:02
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.


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
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Re: M is the sum of the reciprocals of the consecutive integers [#permalink] New post 18 Dec 2012, 21:57
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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 [#permalink] New post 21 Feb 2013, 13:13
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 [#permalink] New post 22 Feb 2013, 01:40
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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RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. NEW!!!

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set. NEW!!!


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Re: M is the sum of the reciprocals of the consecutive integers   [#permalink] 22 Feb 2013, 01:40
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