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 is A
What does inclusive stands for there and what is the best approach to solving this, an Expert pls.
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Inclusive means that it includes the first term (1/201) and the last term (1/300).
The simplest solution to this is to estimate the result. Since there are 100 terms and the middle term is 1/250, the answer is likely to be near 100/250, which reduces to 2/5. That is answer choice (A).
The simplest solution to this is to estimate the result. Since there are 100 terms and the middle term is 1/250, the answer is likely to be near 100/250, which reduces to 2/5. That is answer choice (A).
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We want to find 1/201 + 1/202 + 1/203 + . . . + 1/299 + 1/300Roland2rule 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
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
Brent Hanneson - Creator of GMATPrepNow.com
