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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.

Re: If S is the sum of the reciprocals of the consecutive [#permalink]
25 Nov 2013, 07:12

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Re: If S is the sum of the reciprocals of the consecutive [#permalink]
25 Nov 2013, 07:43

Expert's post

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.

Re: If S is the sum of the reciprocals of the consecutive [#permalink]
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