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The sequence s1, s2, s3,.....sn,...is such that Sn= (1/n) -

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The sequence s1, s2, s3,.....sn,...is such that [#permalink]

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17 Aug 2010, 12:46
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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
[Reveal] Spoiler: OA

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Re: The sequence s1, s2, s3.... [#permalink]

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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]

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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]

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17 Aug 2010, 13:22
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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]

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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!
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The sequence s1, s2, s3,.....sn,...is such that Sn= (1/n) - [#permalink]

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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]

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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.

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Re: Sequences . D.S. 107 , Quant 2 OG [#permalink]

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24 Jan 2011, 02:43
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Merging similar topics.

Also check: knewton-diagnostic-question-106213.html
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Re: DS question : need help [#permalink]

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17 Feb 2011, 10:23
Good Solution Bunuel..
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Re: DS question : need help [#permalink]

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03 Mar 2011, 10:24
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S1=1/1-1/2= 1-1/2
so, Sumk=1-1/k+1
---->k/k+1>9/10
---->k>10
A sufficient
B not sufficient
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Re: DS question : need help [#permalink]

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04 Mar 2011, 12:09
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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

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]

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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.

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Re: The sequence s1, s2, s3.... [#permalink]

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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]

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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]

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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]

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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]

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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]

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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]

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23 Dec 2012, 05:57
Sachin9 wrote:
bunuel,
Please let us know how to solve this one using the A.P formula.
Regards,
Sach

Terms in the sequence are not in AP.
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Re: The sequence s1, s2, s3,.....sn,...is such that Sn= (1/n) - [#permalink]

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