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Reciprocals of consecutive integers

Expert replies
by jdstone550 » Tue Sep 18, 2012 3:54 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

OA = A

Can anyone provide a better explanation than the OG has provided?
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Source: — Problem Solving |

by kevincanspain » Tue Sep 18, 2012 4:14 am
M is the sum of 100 fractions ranging from 1/300 to 1/201. You could reason that the sum of the 100 fractions must not be far from 100/250 =2/5 and thus choose A.

You could also realize that each of the 100 fractions is greater than or equal to 1/300, so their sum must be greater than 100/300 =1/3, leading you to A
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by GMATGuruNY » Tue Sep 18, 2012 4:42 am
jdstone550 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

OA = A

Can anyone provide a better explanation than the OG has provided?
If all of the 100 values were 1/300, the result would be the following sum:
100(1/300) = 1/3.
If all of the 100 values were 1/200, the result would be the following sum:
100(1/200) = 1/2.
There is no value in m less than 1/300 or greater than 1/200.
Thus, the sum of m -- 1/201 + 1/202 +...+ 1/300 -- must be BETWEEN 1/3 and 1/2.

The correct answer is A.
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