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M is the sum of the reciprocals of the consecutive integers from 201

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Re: M is the sum of the reciprocals of the consecutive integers from 201 [#permalink] New post   03 Feb 2020, 22:37
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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


Answer: Option A

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M is the sum of the reciprocals of the consecutive integers from 201 [#permalink] New post   04 May 2021, 02:45
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



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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Re: M is the sum of the reciprocals of the consecutive integers from 201 [#permalink] New post   04 May 2021, 07:10
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sjuniv32 wrote:
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



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.


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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Re: M is the sum of the reciprocals of the consecutive integers from 201 [#permalink] New post   21 May 2021, 13:18
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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Re: M is the sum of the reciprocals of the consecutive integers from 201 [#permalink] New post   27 May 2021, 08:24
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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


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Re: M is the sum of the reciprocals of the consecutive integers from 201 [#permalink] New post   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,
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M is the sum of the reciprocals of the consecutive integers from 201 [#permalink] New post   22 Aug 2021, 23:17
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:
\(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+........+\frac{1}{n} = 2.303 * log(n)\)

(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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Re: M is the sum of the reciprocals of the consecutive integers from 201 [#permalink] New post   12 Oct 2021, 20:35
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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Re: M is the sum of the reciprocals of the consecutive integers from 201 [#permalink] New post   23 Dec 2021, 07:58
fukirua wrote:
Bunuel 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

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


This series is not in AP . hence sum is not required
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Re: M is the sum of the reciprocals of the consecutive integers from 201 [#permalink] New post   19 Jul 2023, 21:36
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 [#permalink] New post   17 Aug 2023, 16:42
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 [#permalink]
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