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If S is the sum of the reciprocals of the consecutive

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If S is the sum of the reciprocals of the consecutive [#permalink] New post 23 Dec 2005, 01:19
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If S is the sum of the reciprocals of the consecutive integers from 91 to 100, inclusive, which of the following is less than S?

I. 1/8
II. 1/9
III. 1/10

A. None
B. I only
C. III only
D. II and III only
E. I, II, and III
[Reveal] Spoiler: OA

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 [#permalink] New post 23 Dec 2005, 01:47
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I think it's C, but I'm not quite sure.

Since we summarize the reciprocals from 100 to 91, we can say also that we add ten numbers who are all (with one exception 1/100) greater than 1/100, so that the sum must be greater than 1/10.

On the other side we can say that we add the reciprocals from 91 to 100, so that the sum has to be less than the sum of ten times 1/91.

We can conclude that the sum has to be less than 1/9 but more than 1/10. That leaves us C as the only possible answer.
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Re: PS: Summation [#permalink] New post 23 Dec 2005, 02:14
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JAI HIND wrote:
If S is the sum of the reciprocals of the consecutive integers from 91 to 100, inclusive, which of the following is less than S?

I. 1/8
II. 1/9
III. 1/10

A. None
B. I only
C. III only
D. II and III only
E. I, II, and III


Answer is C

S = 1/91 + 1/92 + 1/93.....+ 1/100
S = (1/91 + 1/100)(10/2) = 191/1820 = approx 0.1049.....

I. 1/8 = 0.12...
II. 1/9 = 0.11....
III. 1/10 = 0.10
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Re: PS: Summation [#permalink] New post 23 Dec 2005, 04:26
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JAI HIND wrote:
If S is the sum of the reciprocals of the consecutive integers from 91 to 100, inclusive, which of the following is less than S?

I. 1/8
II. 1/9
III. 1/10

A. None
B. I only
C. III only
D. II and III only
E. I, II, and III


From the answer choice we get 1) 1/8 = .125
2) 1/9 = .11 (Approx)
3) 1/10 = .1

1/100 = .01 The other numbers from 1/91 to 1/99 will also be close to .01 but little greater than that . so approx .01 * 10 = .1 So its C.
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 [#permalink] New post 23 Dec 2005, 22:44
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This question is a repetition here.
Anyways lemme explain the solution

1/91+1/92+..........1/100 > 1/100+1/100......10 times
or Summation S >10/100
i.e S>1/10

Similarly
1/91+1/92+..........1/100 < 1/91+1/91......10 times
Summation < 10/91 <1/9

Hence summation is only greater than 1/10.........Hence C
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Re: If S is the sum of the reciprocals of the consecutive [#permalink] New post 25 Nov 2013, 07:12
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Re: If S is the sum of the reciprocals of the consecutive [#permalink] New post 25 Nov 2013, 07:43
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JAI HIND wrote:
If S is the sum of the reciprocals of the consecutive integers from 91 to 100, inclusive, which of the following is less than S?

I. 1/8
II. 1/9
III. 1/10

A. None
B. I only
C. III only
D. II and III only
E. I, II, and III


Given that S=\frac{1}{91}+\frac{1}{92}+\frac{1}{93}+\frac{1}{94}+\frac{1}{95}+\frac{1}{96}+\frac{1}{97}+\frac{1}{98}+\frac{1}{99}+\frac{1}{100}. Notice that 1/91 is the larges term and 1/100 is the smallest term.

If all 10 terms were equal to 1/91, then the sum would be 10/91, but since actual sum is less than that, then we have that S<1/91.

If all 10 terms were equal to 1/100, then the sum would be 10/100=1/10, but since actual sum is more than that, then we have that S>1/10.

Therefore, 1/10 < S < 10/91.

Also, notice that 10/91 < 1/9 < 1/8, thus we have that 1/10 < S < 10/91 < 1/9 < 1/8.

Therefore only 1/10 is less than S.

Answer: C.

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if-k-is-the-sum-of-reciprocals-of-the-consecutive-integers-145365.html
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Re: If S is the sum of the reciprocals of the consecutive [#permalink] New post 10 Jan 2014, 10:40
JAI HIND wrote:
If S is the sum of the reciprocals of the consecutive integers from 91 to 100, inclusive, which of the following is less than S?

I. 1/8
II. 1/9
III. 1/10

A. None
B. I only
C. III only
D. II and III only
E. I, II, and III


We know that there are 10 numbers in the sum: (100-91)+1=10
Take the mean of the sum and times it by 10 to get our sum: (1/ ((100+91)/2)) x 10 = 10/95.5 = 1/9.55

From here we know that the only number which will be smaller than our sum must be divisible by >9.55

Hence only III satisfies. Answer: C
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Re: If S is the sum of the reciprocals of the consecutive [#permalink] New post 20 Feb 2014, 15:30
some person solved this with an Arithmetic progression formula, is that possible, or just a coincidence?
Re: If S is the sum of the reciprocals of the consecutive   [#permalink] 20 Feb 2014, 15:30
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