Question: 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

Correct Answer: A
Solution and Explanation:
Approach Solution 1:

According to the question, ‘M’ is the sum of the reciprocals of the consecutive integers from 201 to 300.
Therefore, we can say,
M=1/201+1/202+1/203+...+1/300.
We must notice here that 1/201 is the largest term and 1/300 is the smallest term.
Now, if we consider, that the sum of all the 100 terms (1/201 + 1/202 + 1/203 + . . . + 1/299 + 1/300) was equal to 1/300, then we would get,
100/300=1/3,

But since the actual sum of these 100 terms (1/201 + 1/202 + 1/203 + . . . + 1/299 + 1/300) would be more than 1/3,
Therefore,
We can say that M > 1/3.

But, if we consider the sum of all the 100 terms (1/201 + 1/202 + 1/203 + . . . + 1/299 + 1/300) to be equal to 1/200,

Then the sum would be
100/200=1/2

But in reality, the sum is less than 1/2.
Therefore, we can say that the value of M lies between 1/3 and 1/2.
Thus, Option A is the correct answer.
1/3 < M < 1/2.

Approach Solution 2

The question is to find the value of M, which is the sum of the reciprocal of values 201 to 300.
We are looking for the value of 1/201 + 1/202 + 1/203 + . . . + 1/299 + 1/300
It is to be noted further, that there are 100 terms in this sum of the fraction.
and
we should also note the highest and the lowest fraction in this sum,
the highest fraction is 1/201,
and the lowest fraction is 1/300.

Now, first, we will consider that the sum of all the fractions (1/201 + 1/202 + 1/203 + . . . + 1/299 + 1/300) will be equal to the smallest fraction of the series (1/201 + 1/202 + 1/203 + . . . + 1/299 + 1/300), which is 1/3.

There we get:

(1/201 + 1/202 + 1/203 + . . . + 1/299 + 1/300) = 100/300 = ⅓
but we know that in reality, the sum of (1/201 + 1/202 + 1/203 + . . . + 1/299 + 1/300) is more than the value of 1/3.

Thus, the value of M is greater than 1/3.
Now we consider that the sum of all the fractions (1/201 + 1/202 + 1/203 + . . . + 1/299 + 1/300) will be equal to the largest fraction of the series (1/201 + 1/202 + 1/203 + . . . + 1/299 + 1/300), which is 1/2.
Thus
(1/201 + 1/202 + 1/203 + . . . + 1/299 + 1/300)=100/200= 1/2.
But, in the original, the value of the sum will be less than ½

Therefore, we can conclude that the value of M will be greater than 1/3, and less than 1/2.
Thus, M will lie in between 1/3 and 1/2.

Hence, Option A 1/3 < M < 1/2, is true among all the options given.

Approach Solution 3:
According to the question, ‘M’ is the sum of the reciprocals of the consecutive integers from 201 to 300.
Therefore, we can say,
M = 1/201+1/202+1/203+...+1/300.

In this series, 1/201+1/202+1/203+...+1/300, the largest fraction is 1/201, as we know when the denominator is less (the least denominator of the series is 201 ), then the value of the fraction is largest,

And when the denominator is more (the highest denominator is 300), then the value of the fraction is the smallest.
Now, if we consider, that the sum of all the 100 terms (1/201 + 1/202 + 1/203 + . . . + 1/299 + 1/300) was equal to 1/300, then we would get,
100/300=1/3,

But since the actual sum of these 100 terms (1/201 + 1/202 + 1/203 + . . . + 1/299 + 1/300) would be more than 1/3,
Therefore,
We can say that M > 1/3.
But, if we consider the sum of all the 100 terms (1/201 + 1/202 + 1/203 + . . . + 1/299 + 1/300) to be equal to 1/200,

Then the sum would be
100/200=1/2

But in reality, the sum is less than 1/2.
Therefore, we can say that the value of M lies between 1/3 and 1/2.
Thus, Option A is the correct answer.
1/3 < M < 1/2.

“M is the sum of the reciprocals of the consecutive integers from”- is a topic of the GMAT Quantitative reasoning section of GMAT. To solve GMAT Problem Solving questions a student must have knowledge about a good amount of qualitative skills. GMAT Quant practice papers improve the mathematical knowledge of the candidates as it represents multiple sorts of quantitative problems.

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