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M is the sum of the reciprocals of the consecutive integers

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by rsarashi » Thu Feb 23, 2017 9:33 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

OAA

What should be the approach.

Thanks.
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by Brent@GMATPrepNow » Thu Feb 23, 2017 9:54 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
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 = A

Cheers,
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by [email protected] » Thu Feb 23, 2017 10:47 am
Hi rsarashi,

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.

Brent's already pointed out the easiest way to figure out the minimum and maximum values of the sum of the reciprocals. You can actually stop working once you figure out the minimum though:

Since 1/300 < 1/201 and the sum of those 100 terms would be 1/3 AT THE MINIMUM, the only answer that's possible would be A. The extra work that Brent did just confirms the maximum value of the sum, but it's 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.

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by Jeff@TargetTestPrep » Wed Mar 01, 2017 5:17 pm
rsarashi 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
Let's first analyze the question. We are trying to find a potential range for M in which M is the sum of the reciprocals from 201 to 300 inclusive. Thus, M is:

1/201 + 1/202 + 1/203 + ... + 1/300

Since we probably would not be expected to do such time-consuming arithmetic (i.e., to add 100 fractions, each with a different denominator), that is exactly why the answer choices are in the form of an inequality. Thus, we do not need to know the EXACT value of M. The easiest way to determine the RANGE of values for M is to use easy numbers that can be quickly manipulated.

Notice that 1/200 is greater than each of the addends and that 1/300 is less than or equal to each of the addends. Therefore, instead of trying to add together 1/201 + 1/202 + 1/203 + ... + 1/300, we are going to add 1/200 one hundred times and 1/300 one hundred times. These two sums will give us a high estimate of M and a low estimate of M. Again, we are adding 1/200 one hundred times and 1/300 one hundred times because there are 100 numbers from 1/201 to 1/300.

Instead of actually adding each one of these values one hundred times, we will simply multiply each value by 100. This gives us:

1/300 x 100 = 1/3

1/200 x 100 = 1/2

We see that M is between 1/3 and 1/2.

Answer: A

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by Matt@VeritasPrep » Thu Mar 02, 2017 5:03 pm
A nice trick when working with unworkable numbers is to find some friendlier numbers with which to compare them. So if we've got, say

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

We know that each of those terms is ≤ 1/2, so we can say

1/2 + 1/2 + 1/2 + 1/2 ≥ 1/2 + 1/3 + 1/4 + 1/5

Here, it's the same idea. We know that

1/200 + 1/200 + ... + 1/200 ≥ 1/200 + 1/201 + ... + 1/300 ≥ 1/300 + 1/300 + ... + 1/300

And since there are 100 terms in each, the 1/200 + 1/200 + ... becomes 100 * (1/200), or 1/2.

Likewise, 1/300 + 1/300 + ... becomes 100 * (1/300), or 1/3. So we're left with

1/2 ≥ 1/201 + 1/202 + ... + 1/300 ≥ 1/3
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