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OG Quant Review Q.107

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OG Quant Review Q.107

by akshatgupta87 » Mon Apr 25, 2011 3:40 am
Q.) The sequence s1,s2,s3,....Sn,... is such that Sn=1/n - 1/(n+1) for all integers n>=1
If k is a positive integer, is the sum of the first k terms of the sequence greater than 9/10?
1) k>10
2) k<19

Someone explain.
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Source: — Data Sufficiency |

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by Anurag@Gurome » Mon Apr 25, 2011 4:14 am
akshatgupta87 wrote:Q.) The sequence s1,s2,s3,....Sn,... is such that Sn=1/n - 1/(n+1) for all integers n>=1
If k is a positive integer, is the sum of the first k terms of the sequence greater than 9/10?
1) k>10
2) k<19

Someone explain.
S1 = 1 - 1/2
S2 = 1/2 - 1/3
S3 = 1/3 - 1/4
...
Sn = 1/n - 1/(n + 1)

Adding, we get, the question as: Is 1 - 1/(n + 1) > 9/10? or 1/(n + 1) < 1/10 or (n + 1) > 10 or n > 9? So, we have check if the Sum > 9.

(1) k > 10, which is obviously SUFFICIENT.

(2) k < 19 implies that the sum may or may not be less than 9. So, (2) is NOT SUFFICIENT.

The correct answer is A.
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by akshatgupta87 » Mon Apr 25, 2011 4:42 am
Can you please explain how do we get the sum of the terms as 1 - 1/(n+1)?
Last edited by akshatgupta87 on Mon Apr 25, 2011 4:48 am, edited 1 time in total.
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by Anurag@Gurome » Mon Apr 25, 2011 4:48 am
akshatgupta87 wrote:Can you please explain how do we get the sum of the terms as 1+ 1/(n+1)?
Please note that we are getting the sum as 1 - 1/(n+1) and not 1+ 1/(n+1).
S1 = 1 - 1/2
S2 = 1/2 - 1/3
S3 = 1/3 - 1/4
...
S(n-1) = 1/(n - 1) - 1/n
Sn = 1/n - 1/(n + 1)

Adding the Left hand and the right hand sides,
S1 + S2 +..._Sn = 1 - 1/(n + 1), because it can be seen that S1 = 1 - 1/2 and S2 = 1/2 - 1/3, so -1/2 and +1/2 cancels, similarly -1/3 and +1/3 cancels....so on...-1/n and +1/n cancels. So, we are just left with the sum = 1 - 1/(n + 1)
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by akshatgupta87 » Mon Apr 25, 2011 5:00 am
Thanks..
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by Anurag@Gurome » Mon Apr 25, 2011 5:01 am
akshatgupta87 wrote:Thanks..
You are welcome.
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