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The sequence s1, s2, s3,.....sn,...is such that [#permalink]
 17 Aug 2010, 12:46
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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? 1) k > 10 2) k < 19
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Re: The sequence s1, s2, s3.... [#permalink]
17 Aug 2010, 13:02
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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
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Re: The sequence s1, s2, s3.... [#permalink]
17 Aug 2010, 13:10
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
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Re: The sequence s1, s2, s3.... [#permalink]
17 Aug 2010, 13:22
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
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Re: The sequence s1, s2, s3.... [#permalink]
17 Aug 2010, 13:35
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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
Cheers!
Ravi
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The sequence s1, s2, s3,.....sn,...is such that Sn= (1/n) - [#permalink]
29 Oct 2010, 20:43
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
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Re: DS question : need help [#permalink]
29 Oct 2010, 21:09
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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. Answer: A.
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Re: Sequences . D.S. 107 , Quant 2 OG [#permalink]
24 Jan 2011, 02:43
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Re: DS question : need help [#permalink]
17 Feb 2011, 10:23
Good Solution Bunuel..
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Re: DS question : need help [#permalink]
03 Mar 2011, 10:24
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Re: DS question : need help [#permalink]
04 Mar 2011, 12:09
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; n=1; 1/1 - (1/1+1) = 1-(1/2) = 1/2 n=2; 1/2 - (1/2+1) = 1/2-1/3 n=3; 1/3 - (1/3+1) = 1/3-1/4 Sum of first 3 = 1/2 + (1/2-1/3) + (1/3 - 1/4) = 1 - 1/4 = First term - last term for sequence Sum of N = 1 - (1/k+1) > 9/10 (k+1)/(k+1)-(1/k+1) > 9/10 k/(k+1) > 9/10 10k>9k+9? k>9? I)sufficient II) sometimes yes sometimes no; insufficient "A"
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Re: DS question : need help [#permalink]
05 Mar 2011, 16:50
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. Answer: A. |
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Re: The sequence s1, s2, s3.... [#permalink]
01 Jan 2012, 08:12
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.
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Re: The sequence s1, s2, s3.... [#permalink]
01 Jan 2012, 09:47
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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.
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Re: The sequence s1, s2, s3.... [#permalink]
02 Jan 2012, 08:27
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. you're right, thanks for clearing this up.
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Re: The sequence s1, s2, s3,.....sn,...is such that [#permalink]
07 Jan 2012, 16:08
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. A.
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Re: The sequence s1, s2, s3,.....sn,...is such that Sn= (1/n) - [#permalink]
29 Mar 2012, 04:57
GMAT Club Legend - awesome work. Really appreciate the amount of effort you put in when laying out your answers... greatly helps the mere mortals!
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Re: The sequence s1, s2, s3,.....sn,...is such that Sn= (1/n) - [#permalink]
22 Dec 2012, 19:44
bunuel, Please let us know how to solve this one using the A.P formula. Regards, Sach
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Re: The sequence s1, s2, s3,.....sn,...is such that Sn= (1/n) - [#permalink]
23 Dec 2012, 05:57
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Re: The sequence s1, s2, s3,.....sn,...is such that Sn= (1/n) -
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23 Dec 2012, 05:57
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