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I used the sum of consecutive numbers formula= first +last*n/2, there are 10 terms in this range.

1/100= 0.01 , 1/91 =slightly >0.01. 91 isnt too far from 100 so it will be slightly greater.

Hence 1/100+1/91 =>0.02 * 5 = slightly >0.1

C- Definitely less than 0.1
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Can you pls explain the sum by mean formula?
Wont it be (1/91 + 1/100)*1/2*10 = (191*10)/(191*100*2) Bunuel
sadovskiya
JAI HIND
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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