Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

Show Tags

17 Aug 2010, 12:46

1

This post received KUDOS

21

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

55% (hard)

Question Stats:

65% (02:39) correct
35% (01:58) wrong based on 405 sessions

HideShow timer Statistics

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) 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

(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."

My Integrated Reasoning Logbook / Diary: http://gmatclub.com/forum/my-ir-logbook-diary-133264.html

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 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]

Show Tags

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

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.

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;

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.

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.

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.
_________________

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]

Show Tags

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.
_________________

I am the master of my fate. I am the captain of my soul. Please consider giving +1 Kudos if deserved!

DS - If negative answer only, still sufficient. No need to find exact solution. PS - Always look at the answers first CR - Read the question stem first, hunt for conclusion SC - Meaning first, Grammar second RC - Mentally connect paragraphs as you proceed. Short = 2min, Long = 3-4 min

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

Show Tags

25 Dec 2013, 11:03

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Best Schools for Young MBA Applicants Deciding when to start applying to business school can be a challenge. Salary increases dramatically after an MBA, but schools tend to prefer...

Marty Cagan is founding partner of the Silicon Valley Product Group, a consulting firm that helps companies with their product strategy. Prior to that he held product roles at...