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M is the sum of the reciprocals of the consecutive integers
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23 May 2020, 01:58
23 May 2020, 01:58
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Question Stats:
78% (01:51) correct
22% (03:10) wrong based on 50 sessions
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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
(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
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23 May 2020, 04:25
23 May 2020, 04:25
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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
Answer: A
Cheers,
Brent
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
Answer: A
Cheers,
Brent
General Discussion
Re: M is the sum of the reciprocals of the consecutive integers
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23 May 2020, 03:31
23 May 2020, 03:31
solution?
