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27 Nov 2019, 23:29
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A
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Difficulty:
25%
(medium)
Question Stats:
75% (01:30) correct
25%
(02:01)
wrong
based on 521
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N represents the sum of reciprocals of consecutive integers from 301 to 400, inclusive. Choose the correct option?
a. 1/4 <N< 1/3
b. 1/6 <N< 1/4
c. 1/8 <N< 1/6
d. 1/10 <N< 1/8
e. 1/12 <N< 1/10
a. 1/4 <N< 1/3
b. 1/6 <N< 1/4
c. 1/8 <N< 1/6
d. 1/10 <N< 1/8
e. 1/12 <N< 1/10
28 Nov 2019, 00:00
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lnm87
Method 1:
Given, \(N = \frac{1}{301} + \frac{1}{302} + \frac{1}{303} + . . . . . . . . . . . . + \frac{1}{400}\)
We Know that, \(\frac{1}{301} < \frac{1}{300}\) and \(\frac{1}{302} < \frac{1}{300}\) so on ...
--> \(N\) is definitely less than \(\frac{1}{300} + \frac{1}{300} + \frac{1}{300} . . . . . . . . . . . . . . + \frac{1}{300}\) (100 times)
--> \(N < 100*\frac{1}{300}\)
--> \(N < \frac{1}{3}\)
We Know that, \(\frac{1}{301} > \frac{1}{400}\) and \(\frac{1}{302} > \frac{1}{400}\) so on ...
--> \(N\) is definitely greater than \(\frac{1}{400} + \frac{1}{400} + \frac{1}{400} . . . . . . . . . . . . . . + \frac{1}{400}\) (100 times)
--> \(N > 100*\frac{1}{400}\)
--> \(N > \frac{1}{4}\)
--> \(\frac{1}{4}\) < N < \(\frac{1}{3}\)
Method 2:
Formula: Sum of 'n' terms of Harmonic Progression is approximately = (number of terms)*(middle term) = \(100*\frac{1}{350} = \frac{1}{3.5}\), which lies between \(\frac{1}{4}\) & \(\frac{1}{3}\)
IMO Option A
27 Nov 2019, 23:36
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