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

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

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The sequence s1, s2, s3, ..., sn, ... is such that \(S_n=\frac{1}{n} - \frac{1}{n+1}\) for all integers \(n\geq{1}\). 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


Data Sufficiency
Question: 107
Category: Arithmetic Sequences
Page: 160
Difficulty: 650


The Official Guide For GMAT® Quantitative Review, 2ND Edition

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

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New post 29 Oct 2010, 20:09
17
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SOLUTION

The sequence s1, s2, s3, ..., sn, ... is such that \(S_n=\frac{1}{n} - \frac{1}{n+1}\) for all integers \(n\geq{1}\). If k is a positive integer, is the sum of the first k terms of the sequence greater than 9/10?

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, ..., sn, ... is such that Sn = 1/n - 1/(n+1)  [#permalink]

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New post 17 Aug 2010, 12:02
9
7
(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
General Discussion
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Re: The sequence s1, s2, s3, ..., sn, ... is such that Sn = 1/n - 1/(n+1)  [#permalink]

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New post 17 Aug 2010, 12:35
2
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 - 1/(n+1)  [#permalink]

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New post 23 Feb 2014, 06:46
5
33

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

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New post 23 Feb 2014, 06:46
15
19
SOLUTION

The sequence s1, s2, s3, ..., sn, ... is such that \(S_n=\frac{1}{n} - \frac{1}{n+1}\) for all integers \(n\geq{1}\). If k is a positive integer, is the sum of the first k terms of the sequence greater than 9/10?

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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Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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

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New post 07 Jul 2014, 08:08
Bunuel wrote:
SOLUTION

The sequence s1, s2, s3, ..., sn, ... is such that \(S_n=\frac{1}{n} - \frac{1}{n+1}\) for all integers \(n\geq{1}\). If k is a positive integer, is the sum of the first k terms of the sequence greater than 9/10?

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.


How exactly did you get sumk = 1- 1/k+1 ? It seems that 1 - 1/4 is not in this form. Maybe if you showed me an example of plugging in I would get it
Math Expert
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Joined: 02 Sep 2009
Posts: 51280
Re: The sequence s1, s2, s3, ..., sn, ... is such that Sn = 1/n - 1/(n+1)  [#permalink]

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New post 07 Jul 2014, 08:14
sagnik2422 wrote:
Bunuel wrote:
SOLUTION

The sequence s1, s2, s3, ..., sn, ... is such that \(S_n=\frac{1}{n} - \frac{1}{n+1}\) for all integers \(n\geq{1}\). If k is a positive integer, is the sum of the first k terms of the sequence greater than 9/10?

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.


How exactly did you get sumk = 1- 1/k+1 ? It seems that 1 - 1/4 is not in this form. Maybe if you showed me an example of plugging in I would get it


Try to work it out yourself.

What is the sum of the first 3 terms?
What is the sum of the first 4 terms?
What is the sum of the first k terms?
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Re: The sequence s1, s2, s3, ..., sn, ... is such that Sn = 1/n - 1/(n+1)  [#permalink]

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New post 27 Nov 2018, 01:30
Bunuel wrote:
The sequence s1, s2, s3, ..., sn, ... is such that \(S_n=\frac{1}{n} - \frac{1}{n+1}\) for all integers \(n\geq{1}\). 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


Data Sufficiency
Question: 107
Category: Arithmetic Sequences
Page: 160
Difficulty: 650


The Official Guide For GMAT® Quantitative Review, 2ND Edition


Similar question to practice: https://gmatclub.com/forum/the-numbers- ... 06213.html
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: The sequence s1, s2, s3, ..., sn, ... is such that Sn = 1/n - 1/(n+1) &nbs [#permalink] 27 Nov 2018, 01:30
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