GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 20 Jan 2019, 19:39

### GMAT Club Daily Prep

#### 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

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

## Events & Promotions

###### Events & Promotions in January
PrevNext
SuMoTuWeThFrSa
303112345
6789101112
13141516171819
20212223242526
272829303112
Open Detailed Calendar
• ### FREE Quant Workshop by e-GMAT!

January 20, 2019

January 20, 2019

07:00 AM PST

07:00 AM PST

Get personalized insights on how to achieve your Target Quant Score.
• ### GMAT Club Tests are Free & Open for Martin Luther King Jr.'s Birthday!

January 21, 2019

January 21, 2019

10:00 PM PST

11:00 PM PST

Mark your calendars - All GMAT Club Tests are free and open January 21st for celebrate Martin Luther King Jr.'s Birthday.

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

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 52296
The sequence s1, s2, s3, ..., sn, ... is such that Sn = 1/n - 1/(n+1)  [#permalink]

### Show Tags

17 Aug 2010, 11:46
1
44
00:00

Difficulty:

65% (hard)

Question Stats:

66% (02:02) correct 34% (02:16) wrong based on 881 sessions

### HideShow timer Statistics

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

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 52296
The sequence s1, s2, s3, ..., sn, ... is such that Sn = 1/n - 1/(n+1)  [#permalink]

### Show Tags

29 Oct 2010, 20:09
17
27
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.

_________________
Manager
Joined: 20 Jul 2010
Posts: 62
Re: The sequence s1, s2, s3, ..., sn, ... is such that Sn = 1/n - 1/(n+1)  [#permalink]

### Show Tags

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
Manager
Joined: 20 Jul 2010
Posts: 62
Re: The sequence s1, s2, s3, ..., sn, ... is such that Sn = 1/n - 1/(n+1)  [#permalink]

### Show Tags

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
Math Expert
Joined: 02 Sep 2009
Posts: 52296
The sequence s1, s2, s3, ..., sn, ... is such that Sn = 1/n - 1/(n+1)  [#permalink]

### Show Tags

23 Feb 2014, 06:46
5
33
Math Expert
Joined: 02 Sep 2009
Posts: 52296
Re: The sequence s1, s2, s3, ..., sn, ... is such that Sn = 1/n - 1/(n+1)  [#permalink]

### Show Tags

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.

_________________
Intern
Joined: 20 May 2014
Posts: 32
Re: The sequence s1, s2, s3, ..., sn, ... is such that Sn = 1/n - 1/(n+1)  [#permalink]

### Show Tags

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.

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
Joined: 02 Sep 2009
Posts: 52296
Re: The sequence s1, s2, s3, ..., sn, ... is such that Sn = 1/n - 1/(n+1)  [#permalink]

### Show Tags

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.

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?
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 52296
Re: The sequence s1, s2, s3, ..., sn, ... is such that Sn = 1/n - 1/(n+1)  [#permalink]

### Show Tags

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

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

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.