Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Game of Timers is a team-based competition based on solving GMAT questions to win epic prizes! Starting July 1st, compete to win prep materials while studying for GMAT! Registration is Open!

Re: M is the sum of the reciprocals of the consecutive integers from 201
[#permalink]

Show Tags

04 Sep 2016, 20:55

Hi Manonamission,

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. You can actually stop working once you figure out the minimum value...

Since 1/300 < 1/201 and the sum of those 100 terms would be (100)(1/300) = 1/3 AT THE MINIMUM, the only answer that's possible would be A. The extra work that you could do to find the maximum value just confirms what the sum would be at that 'level', but but that work is 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.

M is the sum of the reciprocals of the consecutive integers from 201
[#permalink]

Show Tags

26 Nov 2018, 06:32

mike34170 wrote:

I'd handle this differently:

Numbers range from 201 to 300, so the average is approximatively 250.

Then, the sum of all reciprocals equals to 100 times 1/250, i.e. 100/250 ---> 1/2,5

Answer A

Although you do get the answer, it is just a coincidence as the correct way to solve is given above by bunel. Let me explain why this doesn't work using a simpler example!

Let's say we have the following sum:

1/1 + 1/2 + 1/3 + 1/4 + 1/5.

If we solve this with a calculator, we get 2.283

If we use the approach you suggested for this one, we would have the middle value (1/3) multiplied by 5.

(1/3) * 5 = 5/3 = 1.66

This is not the same as 2.283. So the method doesn't really work when we have fractions.

If we have a sum that doesn't involve fractions, the method does work. For example:

Re: M is the sum of the reciprocals of the consecutive integers from 201
[#permalink]

Show Tags

25 Mar 2019, 09:17

Top Contributor

Walkabout 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

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