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555-605 (Medium)|   Arithmetic|                              
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BrentGMATPrepNow
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M is the sum of the reciprocals of the consecutive integers from 201 04 May 2021, 07:10
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sjuniv32
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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


BrentGMATPrepNow, could you please explain this approach too? I didn't catch how 100/250 ---> 1/2.5 helps to reach answer option A. Thanks in advance.
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The idea here is that the average of all 100 numbers (from 201 to 300) is approximately 250.
So the solution replaces all 100 numbers with 250's {250, 250, 250, 250, 250, ......250, 250, 250}
So the sum of the 100 reciprocals = 1/250 + 1/250 + 1/250 + 1/250 + 1/250 + .....+ 1/250 + 1/250
= 100/250
= 0.4

Since 1/5 < 0.4 < 1/3, the correct answer is B
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M is the sum of the reciprocals of the consecutive integers from 201 27 May 2021, 08:24
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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
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Answer: Option A

Video solution by GMATinsight

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M is the sum of the reciprocals of the consecutive integers from 201 19 Aug 2021, 02:20
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M = \(\frac{1}{201} + \frac{1}{202} +...... + \frac{1}{300} \)

There are a total of 100 terms in this series.
When you compare each term in the series , the minimum value in the series is 1/300

Since there are 100 terms in the series and each term is greater than 1/300. We can conclude that the sum M should greater than\( 100* \frac{1}{300}\).

M > 1/3

When you analyze the option, we can see that they are arranged in descending order. In option A, the lower limit of the range is 1/3
while in option B , the upper limit is 1/3. Similarly as the other options are also in the decreasing order of the range, So Option B,C,D,E can be blindly eliminated as Sum should be greater than 1/3

Thanks,
Clifin J Francis,
GMAT SME
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M is the sum of the reciprocals of the consecutive integers from 201 28 Mar 2025, 04:11
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can i not simplify 1/201 and 1/300 to : 1/2 and 1/3 respectively intuitively ? the ranges will be the same, I don't care about absolute values, right?

and do N = 1/2 - 1/3 + 1 = 7/6
AVG = ( 1/3 + 1/2 ) / 2 = 5/12 ---- 35/72 < 1/2 .. so A only one that satisfies

Bunuel
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

Given that \(M=\frac{1}{201}+\frac{1}{202}+\frac{1}{203}+...+\frac{1}{300}\). Notice that 1/201 is the largest term and 1/300 is the smallest term.

If all 100 terms were equal to 1/300, then the sum would be 100/300 = 1/3, but since the actual sum is more than that, we have that M > 1/3.

If all 100 terms were equal to 1/200, then the sum would be 100/200 = 1/2, but since the actual sum is less than that, we have that M < 1/2.

Therefore, 1/3 < M < 1/2.

Answer: A.
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M is the sum of the reciprocals of the consecutive integers from 201 28 Mar 2025, 04:35
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INprimesItrust
can i not simplify 1/201 and 1/300 to : 1/2 and 1/3 respectively intuitively ? the ranges will be the same, I don't care about absolute values, right?

and do N = 1/2 - 1/3 + 1 = 7/6
AVG = ( 1/3 + 1/2 ) / 2 = 5/12 ---- 35/72 < 1/2 .. so A only one that satisfies

Bunuel
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

Given that \(M=\frac{1}{201}+\frac{1}{202}+\frac{1}{203}+...+\frac{1}{300}\). Notice that 1/201 is the largest term and 1/300 is the smallest term.

If all 100 terms were equal to 1/300, then the sum would be 100/300 = 1/3, but since the actual sum is more than that, we have that M > 1/3.

If all 100 terms were equal to 1/200, then the sum would be 100/200 = 1/2, but since the actual sum is less than that, we have that M < 1/2.

Therefore, 1/3 < M < 1/2.

Answer: A.
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Unless I’m missing something in your logic, that approach doesn’t seem correct.

1/201 is about 100 times smaller than 1/2, and 1/300 is exactly 100 times smaller than 1/3—so simplifying to 1/2 and 1/3 loses too much precision for a tight range like this.

Also, not sure what you’re doing with 1/2 - 1/3 + 1 = 7/6, or why you’re averaging 1/2 and 1/3—it doesn’t reflect the actual sum or its bounds.

Just FYI: 1/201 + 1/202 + ... + 1/300 ≈ 0.4.
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M is the sum of the reciprocals of the consecutive integers from 201 04 Mar 2026, 12:06
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Walkabout
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
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A




Nick Slavkovich, GMAT/GRE tutor with 20+ years of experience

[email protected]
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M is the sum of the reciprocals of the consecutive integers from 201 03 May 2026, 09:21
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I solved this under 2 mins and here is how i approached this question-

I converted the available options in decimal terms and got an estimation of the ranges of the options.
1. 0.33<m<0.5
2. 0.2<m<0.33
3. 0.14<m<0.2
4. 0.11<m<0.14
5. 0.08<m<0.11

Now looking back at the question:

{1/201, 1/202, 1/203, ..., 1/300} I got a rough enough estimate by solving for 1, 2 values as to which is the range of sum I am looking for.

Answer- A
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