It is currently 17 Nov 2017, 11:09

### 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 June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

Retired Moderator
Status: 2000 posts! I don't know whether I should feel great or sad about it! LOL
Joined: 04 Oct 2009
Posts: 1629

Kudos [?]: 1120 [1], given: 109

Location: Peru
Schools: Harvard, Stanford, Wharton, MIT & HKS (Government)
WE 1: Economic research
WE 2: Banking
WE 3: Government: Foreign Trade and SMEs
The sequence s1, s2, s3, ..., sn, ... is such that Sn= 1/n - 1/(n + 1) [#permalink]

### Show Tags

17 Aug 2010, 12:46
1
KUDOS
26
This post was
BOOKMARKED
00:00

Difficulty:

55% (hard)

Question Stats:

68% (01:37) correct 32% (01:48) wrong based on 473 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
[Reveal] Spoiler: OA

_________________

"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

GMAT Club Premium Membership - big benefits and savings

Kudos [?]: 1120 [1], given: 109

Manager
Joined: 20 Jul 2010
Posts: 77

Kudos [?]: 77 [6], given: 32

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

### Show Tags

17 Aug 2010, 13:02
6
KUDOS
6
This post was
BOOKMARKED
(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

Kudos [?]: 77 [6], given: 32

Director
Status: Apply - Last Chance
Affiliations: IIT, Purdue, PhD, TauBetaPi
Joined: 18 Jul 2010
Posts: 681

Kudos [?]: 169 [0], given: 15

Schools: Wharton, Sloan, Chicago, Haas
WE 1: 8 years in Oil&Gas
Re: The sequence s1, s2, s3, ..., sn, ... is such that Sn= 1/n - 1/(n + 1) [#permalink]

### Show Tags

17 Aug 2010, 13:22
1
This post was
BOOKMARKED
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
_________________

Consider kudos, they are good for health

Kudos [?]: 169 [0], given: 15

Manager
Joined: 20 Jul 2010
Posts: 77

Kudos [?]: 77 [2], given: 32

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

### Show Tags

17 Aug 2010, 13:35
2
KUDOS
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

Kudos [?]: 77 [2], given: 32

Math Expert
Joined: 02 Sep 2009
Posts: 42249

Kudos [?]: 132500 [12], given: 12323

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

### Show Tags

29 Oct 2010, 21:09
12
KUDOS
Expert's post
24
This post was
BOOKMARKED
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.

_________________

Kudos [?]: 132500 [12], given: 12323

Math Expert
Joined: 02 Sep 2009
Posts: 42249

Kudos [?]: 132500 [0], given: 12323

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

### Show Tags

24 Jan 2011, 02:43
Expert's post
1
This post was
BOOKMARKED
Similar question: https://gmatclub.com/forum/the-numbers- ... 06213.html
_________________

Kudos [?]: 132500 [0], given: 12323

Manager
Status: I am Midnight's Child !
Joined: 04 Dec 2009
Posts: 140

Kudos [?]: 83 [0], given: 11

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

### Show Tags

17 Feb 2011, 10:23
Good Solution Bunuel..
_________________

Argument : If you love long trips, you love the GMAT.
Conclusion : GMAT is long journey.

What does the author assume ?
Assumption : A long journey is a long trip.

GMAT Club Premium Membership - big benefits and savings

Kudos [?]: 83 [0], given: 11

Director
Status: No dream is too large, no dreamer is too small
Joined: 14 Jul 2010
Posts: 609

Kudos [?]: 1153 [1], given: 39

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

### Show Tags

03 Mar 2011, 10:24
1
KUDOS
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
_________________

Collections:-
PSof OG solved by GC members: http://gmatclub.com/forum/collection-ps-with-solution-from-gmatclub-110005.html
DS of OG solved by GC members: http://gmatclub.com/forum/collection-ds-with-solution-from-gmatclub-110004.html
100 GMAT PREP Quantitative collection http://gmatclub.com/forum/gmat-prep-problem-collections-114358.html
Collections of work/rate problems with solutions http://gmatclub.com/forum/collections-of-work-rate-problem-with-solutions-118919.html
Mixture problems in a file with best solutions: http://gmatclub.com/forum/mixture-problems-with-best-and-easy-solutions-all-together-124644.html

Kudos [?]: 1153 [1], given: 39

Manager
Status: TIME FOR 700+
Joined: 06 Dec 2010
Posts: 202

Kudos [?]: 49 [0], given: 55

Schools: Fuqua
WE 1: Research in Neurology
WE 2: MORE research in Neurology
Re: The sequence s1, s2, s3, ..., sn, ... is such that Sn= 1/n - 1/(n + 1) [#permalink]

### Show Tags

04 Mar 2011, 12:09
1
This post was
BOOKMARKED
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"
_________________

Back to the grind, goal 700+

Kudos [?]: 49 [0], given: 55

Director
Joined: 01 Feb 2011
Posts: 726

Kudos [?]: 146 [0], given: 42

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

### Show Tags

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.

Kudos [?]: 146 [0], given: 42

Manager
Joined: 12 Oct 2011
Posts: 128

Kudos [?]: 254 [0], given: 23

GMAT 1: 700 Q48 V37
GMAT 2: 720 Q48 V40
Re: The sequence s1, s2, s3, ..., sn, ... is such that Sn= 1/n - 1/(n + 1) [#permalink]

### Show Tags

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.

Kudos [?]: 254 [0], given: 23

CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2757

Kudos [?]: 1907 [1], given: 235

Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Re: The sequence s1, s2, s3, ..., sn, ... is such that Sn= 1/n - 1/(n + 1) [#permalink]

### Show Tags

01 Jan 2012, 09:47
1
KUDOS
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.
_________________

Fight for your dreams :For all those who fear from Verbal- lets give it a fight

Money Saved is the Money Earned

Jo Bole So Nihaal , Sat Shri Akaal

GMAT Club Premium Membership - big benefits and savings

Gmat test review :
http://gmatclub.com/forum/670-to-710-a-long-journey-without-destination-still-happy-141642.html

Kudos [?]: 1907 [1], given: 235

Manager
Joined: 12 Oct 2011
Posts: 128

Kudos [?]: 254 [0], given: 23

GMAT 1: 700 Q48 V37
GMAT 2: 720 Q48 V40
Re: The sequence s1, s2, s3, ..., sn, ... is such that Sn= 1/n - 1/(n + 1) [#permalink]

### Show Tags

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.

Kudos [?]: 254 [0], given: 23

Manager
Joined: 29 Jul 2011
Posts: 104

Kudos [?]: 72 [0], given: 6

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

Kudos [?]: 72 [0], given: 6

Intern
Joined: 26 Feb 2012
Posts: 6

Kudos [?]: 1 [0], given: 3

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

### Show Tags

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!

Kudos [?]: 1 [0], given: 3

Director
Status: Gonna rock this time!!!
Joined: 22 Jul 2012
Posts: 506

Kudos [?]: 72 [0], given: 562

Location: India
GMAT 1: 640 Q43 V34
GMAT 2: 630 Q47 V29
WE: Information Technology (Computer Software)
Re: The sequence s1, s2, s3, ..., sn, ... is such that Sn= 1/n - 1/(n + 1) [#permalink]

### Show Tags

22 Dec 2012, 19:44
bunuel,
Please let us know how to solve this one using the A.P formula.
Regards,
Sach
_________________

hope is a good thing, maybe the best of things. And no good thing ever dies.

Who says you need a 700 ?Check this out : http://gmatclub.com/forum/who-says-you-need-a-149706.html#p1201595

My GMAT Journey : http://gmatclub.com/forum/end-of-my-gmat-journey-149328.html#p1197992

Kudos [?]: 72 [0], given: 562

Math Expert
Joined: 02 Sep 2009
Posts: 42249

Kudos [?]: 132500 [0], given: 12323

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

### Show Tags

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.
_________________

Kudos [?]: 132500 [0], given: 12323

Intern
Joined: 20 Jan 2016
Posts: 14

Kudos [?]: 1 [0], given: 34

Schools: HBS '18
Re: The sequence s1, s2, s3, ..., sn, ... is such that Sn= 1/n - 1/(n + 1) [#permalink]

### Show Tags

17 Oct 2016, 17:02
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.

Thanks. What are some similar questions to practice?

Kudos [?]: 1 [0], given: 34

Math Expert
Joined: 02 Sep 2009
Posts: 42249

Kudos [?]: 132500 [0], given: 12323

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

### Show Tags

18 Oct 2016, 02:00
oasis90 wrote:
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.

Thanks. What are some similar questions to practice?

Check DS and PS sequence questions from our questions bank:
search.php?search_id=tag&tag_id=111
search.php?search_id=tag&tag_id=112

Hope it helps.
_________________

Kudos [?]: 132500 [0], given: 12323

Manager
Joined: 23 Dec 2013
Posts: 235

Kudos [?]: 14 [0], given: 21

Location: United States (CA)
GMAT 1: 760 Q49 V44
GPA: 3.76
Re: The sequence s1, s2, s3, ..., sn, ... is such that Sn= 1/n - 1/(n + 1) [#permalink]

### Show Tags

03 Jun 2017, 21:04
metallicafan 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

This problem is best tackled by identifying the first term.

In this case, S1 = 1/1 - 1/2.

S2 = 1/2 - 1/3

S3 = 1/3 - 1/4

If you add them together, all but the first and last terms cancel. So to determine if the first k terms are greater than 9/10, we need to determine if 1 - 1/(1+k) > 9/10.

In other words, is 1/(1+k) < 1/10?

10 < 1+k?

9<k?

Statement 1) K > 10. Sufficient.

Statement 2) K < 19. Not sufficient.

Kudos [?]: 14 [0], given: 21

Re: The sequence s1, s2, s3, ..., sn, ... is such that Sn= 1/n - 1/(n + 1)   [#permalink] 03 Jun 2017, 21:04

Go to page    1   2    Next  [ 22 posts ]

Display posts from previous: Sort by