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The sequence s1, s2, s3,.....sn,...is such that [#permalink]
17 Aug 2010, 11:46
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Difficulty:
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Question Stats:
63% (02:38) correct
37% (02:12) wrong based on 239 sessions
Please your help:
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 \(9/10\)?
Re: The sequence s1, s2, s3.... [#permalink]
17 Aug 2010, 12:02
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(1) Take K as 11. So, Sum = S1 + S2 + S3 + S4 + S5+ S6 + S7 + S8 + S9 + S10 + S11 Where, S1 = 1 - (1/2) S2 = (1/2) - (1/3) S3 = (1/3) - (1/4) ... S11 = (1/11) - (1/12)
Which implies, Sum = 1 - (1/12) /* The terms like +1/2, -1/2, +1/3, -1/3 will be added to zero. Only the first and last numbers remains */ ==> Sum = 1 - 0.0XXXX > 9/10
If you take k as 12, the SUM = 1 - (1/13) which is again > 0.9 Hence SUFFICIENT
(2) K < 19 Consider K as 2. Then the sum is = 1 - (1/2) + (1/2) - (1/3) = 1 - (1/3) which is Less than 9/10 Consider K as 11, Then the sum is greater than 9/10 /* We already proved this in (1) above */ Hence (2) is In Sufficient
Re: The sequence s1, s2, s3.... [#permalink]
17 Aug 2010, 12:10
nravi4 wrote:
(1) Take K as 11. So, Sum = S1 + S2 + S3 + S4 + S5+ S6 + S7 + S8 + S9 + S10 + S11 Where, S1 = 1 - (1/2) S2 = (1/2) - (1/3) S3 = (1/3) - (1/4) ... S11 = (1/11) - (1/12)
Which implies, Sum = 1 - (1/12) /* The terms like +1/2, -1/2, +1/3, -1/3 will be added to zero. Only the first and last numbers remains */ ==> Sum = 1 - 0.0XXXX > 9/10
If you take k as 12, the SUM = 1 - (1/13) which is again > 0.9 Hence SUFFICIENT
(2) K < 19 Consider K as 2. Then the sum is = 1 - (1/2) + (1/2) - (1/3) = 1 - (1/3) which is Less than 9/10 Consider K as 11, Then the sum is greater than 9/10 /* We already proved this in (1) above */ Hence (2) is In Sufficient
Wow, you are good! I have a question, how did you know that you had to test some numbers in the sequence?, Why in that way? I tried to test them but my way was more complex and difficult :s _________________
"Life’s battle doesn’t always go to stronger or faster men; but sooner or later the man who wins is the one who thinks he can."
Re: The sequence s1, s2, s3.... [#permalink]
17 Aug 2010, 12:22
If you notice, the sum is simply given by Sn = 1 -1/(n+1) = n/(n+1). hence S9 = 9/10. Adding further terms only increases the sum - hence S10 > S9 etc... 1) k>10, clearly sufficient as S10>9/10 = S9 2) k<19, cant say much K can be 1 = 1/2<9/10 but K=11 makes 11/12>9/10 Hence A _________________
Re: The sequence s1, s2, s3.... [#permalink]
17 Aug 2010, 12:35
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I am not a master in choosing the right number, but as i read & practising the same ... Choose the numbers which are closer to the lowest range & Highest range first.
Example: (1) K > 10. The lowest range number here is: 11. If 11 is sufficient then take 12. If 12 is also sufficient find out if you can make a generalized statement as Sufficient?
The highest range number is: Infinity
(2) K < 19 The lowest range number (according to problem specificaitons) is: 1 The highest range number is: 18
The sequence s1, s2, s3,.....sn,...is such that Sn= (1/n) - [#permalink]
29 Oct 2010, 19:43
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
Re: DS question : need help [#permalink]
29 Oct 2010, 20:09
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satishreddy wrote:
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
Given: \(s_n=\frac{1}{n}-\frac{1}{n+1}\) for \(n\geq{1}\). So: \(s_1=1-\frac{1}{2}\); \(s_2=\frac{1}{2}-\frac{1}{3}\); \(s_3=\frac{1}{3}-\frac{1}{4}\); ...
If you sum the above 3 terms you'll get: \(s_1+s_2+s_3=(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})=1-\frac{1}{4}\) (everything but the first and the last numbers will cancel out). So the sum of first \(k\) terms is fgiven by the formula \(sum_k=1-\frac{1}{k+1}\).
Question: is \(sum_k=1-\frac{1}{k+1}>\frac{9}{10}\)? --> is \(\frac{k}{k+1}>\frac{9}{10}\)? --> is \(k>9\)?
(1) k > 10. Sufficient. (2) k < 19. Not sufficient.
Re: DS question : need help [#permalink]
04 Mar 2011, 11:09
satishreddy wrote:
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
I) (1/n) - (1/n+1) where N>=1; k = sum of the sequence;
Re: DS question : need help [#permalink]
05 Mar 2011, 15:50
Good Question whoever posted it.
Great solution by Bunuel. Good that you didn't solve the sn formula initially to 1/n(n+1) . that way its easy to cancel out terms.
Bunuel wrote:
satishreddy wrote:
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
Given: \(s_n=\frac{1}{n}-\frac{1}{n+1}\) for \(n\geq{1}\). So: \(s_1=1-\frac{1}{2}\); \(s_2=\frac{1}{2}-\frac{1}{3}\); \(s_3=\frac{1}{3}-\frac{1}{4}\); ...
If you sum the above 3 terms you'll get: \(s_1+s_2+s_3=(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})=1-\frac{1}{4}\) (everything but the first and the last numbers will cancel out). So the sum of first \(k\) terms is fgiven by the formula \(sum_k=1-\frac{1}{k+1}\).
Question: is \(sum_k=1-\frac{1}{k+1}>\frac{9}{10}\)? --> is \(\frac{k}{k+1}>\frac{9}{10}\)? --> is \(k>9\)?
(1) k > 10. Sufficient. (2) k < 19. Not sufficient.
Re: The sequence s1, s2, s3.... [#permalink]
01 Jan 2012, 07:12
mainhoon wrote:
If you notice, the sum is simply given by Sn = 1 -1/(n+1) = n/(n+1). hence S9 = 9/10. Adding further terms only increases the sum - hence S10 > S9 etc... 1) k>10, clearly sufficient as S10>9/10 = S9 2) k<19, cant say much K can be 1 = 1/2<9/10 but K=11 makes 11/12>9/10 Hence A
I don't think sn=1/n-1/(n+1) is equal to n/(n+1), it's equal to 1/n(n+1).
I find that sequence problems come down to pattern recognition, nravi's solution is probably what it's supposed to look like, you have to see that the terms other than 1 and 1/12 cancel each other out.
Re: The sequence s1, s2, s3.... [#permalink]
01 Jan 2012, 08:47
1
This post received KUDOS
BN1989 wrote:
mainhoon wrote:
If you notice, the sum is simply given by Sn = 1 -1/(n+1) = n/(n+1). hence S9 = 9/10. Adding further terms only increases the sum - hence S10 > S9 etc... 1) k>10, clearly sufficient as S10>9/10 = S9 2) k<19, cant say much K can be 1 = 1/2<9/10 but K=11 makes 11/12>9/10 Hence A
I don't think sn=1/n-1/(n+1) is equal to n/(n+1), it's equal to 1/n(n+1).
I find that sequence problems come down to pattern recognition, nravi's solution is probably what it's supposed to look like, you have to see that the terms other than 1 and 1/12 cancel each other out.
sum of \([\frac{1}{n} - \frac{1}{(n+1)}]\) from = 1 to n=n => \(\frac{n}{(n+1)}\) \(1-\frac{1}{2} + \frac{1}{2}-\frac{1}{3} + \frac{1}{3}-\frac{1}{4} + ............... \frac{1}{n} -\frac{1}{(n+1)}\) = \(1 - \frac{1}{(n+1)}\) =\(\frac{n}{(n+1)}\) because all the terms between 1st and last are cancelled. _________________
Re: The sequence s1, s2, s3.... [#permalink]
02 Jan 2012, 07:27
gurpreetsingh wrote:
BN1989 wrote:
mainhoon wrote:
If you notice, the sum is simply given by Sn = 1 -1/(n+1) = n/(n+1). hence S9 = 9/10. Adding further terms only increases the sum - hence S10 > S9 etc... 1) k>10, clearly sufficient as S10>9/10 = S9 2) k<19, cant say much K can be 1 = 1/2<9/10 but K=11 makes 11/12>9/10 Hence A
I don't think sn=1/n-1/(n+1) is equal to n/(n+1), it's equal to 1/n(n+1).
I find that sequence problems come down to pattern recognition, nravi's solution is probably what it's supposed to look like, you have to see that the terms other than 1 and 1/12 cancel each other out.
sum of \([\frac{1}{n} - \frac{1}{(n+1)}]\) from = 1 to n=n => \(\frac{n}{(n+1)}\) \(1-\frac{1}{2} + \frac{1}{2}-\frac{1}{3} + \frac{1}{3}-\frac{1}{4} + ............... \frac{1}{n} -\frac{1}{(n+1)}\) = \(1 - \frac{1}{(n+1)}\) =\(\frac{n}{(n+1)}\) because all the terms between 1st and last are cancelled.
Re: The sequence s1, s2, s3,.....sn,...is such that [#permalink]
07 Jan 2012, 15:08
Tricky one, needs so serious rephrasing of the stem:
Rephrase: for the k number of elements, the sum of all the terms will be = 1/x - 1/(x+k) (try yourself! - intermediate terms cancel out)
1. Say k = 11, we get 1/1 - 1/(1+11) = 11/12 > 9/10. Suff 2. k<19. it could be k=11 as above or k=8, where you get 1/1 - 1/(1+8) = 8/9 < 9/10. Insuff.
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Re: The sequence s1, s2, s3,.....sn,...is such that Sn= (1/n) - [#permalink]
25 Dec 2013, 10:03
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