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555-605 Level|   Arithmetic|                              
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GMATinsight
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M is the sum of the reciprocals of the consecutive integers from 201 03 Feb 2020, 22:37
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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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Answer: Option A

Find soltion attached
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Screenshot 2020-02-04 at 11.05.40 AM.png
Screenshot 2020-02-04 at 11.05.40 AM.png [ 368.97 KiB | Viewed 7618 times ]

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M is the sum of the reciprocals of the consecutive integers from 201 04 May 2021, 02:45
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mike34170
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
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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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M is the sum of the reciprocals of the consecutive integers from 201 04 May 2021, 07:10
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sjuniv32
mike34170
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 21 May 2021, 13:18
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1/201 is approximately: 0.005
1/300 is approximately: 0.003
avg is : 0.004

since there are 100 elements; so sum = 0.4

only option A suits
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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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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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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 22 Aug 2021, 23:17
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My approach is nowhere close to Bunuel's solution above, still putting this out, in case anybody finds it useful.

We have a result:

Quote:
\(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+........+\frac{1}{n} = 2.303 * log(n)\)
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(PS : I dont have the derivation, you can just remember it, if you want to)

For this question, we need to calculate
\((1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{300}) - (1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200})\)
\(= 2.303 log(300) - 2.303 log(200)\)
\(=2.303 log (3/2)\)
\(=2.303 (log(3) - log(2))\)
\(=2.303 (0.4771-0.3010)\)
\(=2.3*0.17 (approx)\)
<= 0.4

(A) satisfies
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M is the sum of the reciprocals of the consecutive integers from 201 12 Oct 2021, 20:35
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This is a longer method, but the one I used to solve in 2:14

You can pair the reciprocals:

1/200 + 1/300
3/600 + 2/600 = 5/600
5/600 = 1/120
1/120 * 50 = 50/120
5/12

1/3 < 5/12 < 1/2
****4/12 < 5/12 < 6/12****

Note: in reality, the true answer is just BARELY shy of 5/12 since 1/200 not 1/201 was used to get a common deonominator.
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M is the sum of the reciprocals of the consecutive integers from 201 23 Dec 2021, 07:58
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fukirua
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 larges 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 actual sum is more than that, then 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 actual sum is less than that, then we have that M<1/2.

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

Answer: A.

WOW, that is elegant!
Bunuel please suggest if we can use the next assumption:
Arithmetic mean of the elements a1 and a100= 501/(300*2*201). Sum of all elements = 100*Arithmetic mean=167/(6*67)= 167/402 , which is definitely more than 1/3. A
Thanks
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This series is not in AP . hence sum is not required
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M is the sum of the reciprocals of the consecutive integers from 201 19 Jul 2023, 21:36
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Use estimation here, Since the sum of first and last is around 1/2.5, and all are consecutive. A is the closest match.
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M is the sum of the reciprocals of the consecutive integers from 201 17 Aug 2023, 16:42
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I don't know if the method is right, but I thought of Σχ = average * N

Finding the average since they are consecutive: ( [1][/201] + [1][/300] ) / 2 = [501][/201*300*2] -> Average
Finding the number of multiples N: (301-200) + 1 (since it says inclusive) = 100 ->N
Σχ = [501][/201*300*2] * 100 -----> approx. [500][/200 * 300*2] *100 = 5/12 = .42
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M is the sum of the reciprocals of the consecutive integers from 201 18 Aug 2024, 11:20
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Very easy question, but the condition here is thinking clearly with patience, or else you'll get confused really fast.

Posted from my mobile device
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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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