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i used the sum of a series in Ap approach which I found to be more simple
for 5 terms in ap it gives =
5/2(1/16+1/20)
(5/2)*(9/80)
(9/32)
which is closest to 1/4
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Bunuel
The sum of the reciprocals of the 5 consecutive integers from 16 to 20 is closest to which of the following ?

A. 1/2
B. 1/3
C. 1/4
D. 1/5
E. 1/6
­i just approached this Q in a diff manner , 1/20 is 0.05 and 1/16 is 0.06 (wont take long to figure that out ) from there i just added 0.05 thrice and 0.06 twice giving me 0.27 hence 1/4 . Is this alright to do ?

thanks 
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The middle value is 1/18.

Therefore take 5/18 since there are 5 terms. This is equal to 1/3.6 which is approximatley 1/4.

Is this approach applicable?
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Not sure whether it's the correct approach, but

I'd take 1/20 = 0.05, now all other reciprocals will be equal almost close to 0.05, hence 0.05*5 ~ 0.25

Hence, answer is C
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\(5/16>n>5/20 0.31>n>0.25\\
\\
1/16+1/17+1/18+1/19+1/20 =ln(20/16)\\
=2.3(log(5)-2log(2))=2.3(0.7-0.6)=2.3*0.1\\
=0.23\\
\\
\\
\\
\\
log 2=0.3, log 5=0.7­\)­

1/4 is closest
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Asked: The sum of the reciprocals of the 5 consecutive integers from 16 to 20 is closest to which of the following ?

The sum of the reciprocals of the 5 consecutive integers from 16 to 20 \(= 1/16 + 1/17 + 1/18 + 1/19 + 1/20\)

\(5*1/15 = 1/3 > 1/16 + 1/17 + 1/18 + 1/19 + 1/20 > 5*1/20 = 1/4\)

Since all reciprocals are nearer to 1/20, therefore, their sum needs to be closest to 1/4

​​​​​​​IMO C
­
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1/16 + 1/17 + 1/18 + 1/19 + 1/20 is a harmonic progression.

The sum of a harmonic progression is -
Number of terms * middle term

Which in this case is 5 * 1/18 = 5/18 = 0.278

which is closest to 1/4

Hope this helps :)
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