Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 26 May 2017, 22:40

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

In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk

Author Message
TAGS:

Hide Tags

Director
Joined: 28 Jul 2011
Posts: 549
Location: United States
GPA: 3.86
WE: Accounting (Commercial Banking)
Followers: 3

Kudos [?]: 238 [3] , given: 16

In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk [#permalink]

Show Tags

26 Jan 2012, 04:15
3
KUDOS
15
This post was
BOOKMARKED
00:00

Difficulty:

35% (medium)

Question Stats:

68% (02:37) correct 32% (01:43) wrong based on 587 sessions

HideShow timer Statistics

In the sequence $$x_0, \ x_1, \ x_2, \ ... \ x_n$$, each term from $$x_1$$ to $$x_k$$ is 3 greater than the previous term, and each term from $$x_{k+1}$$ to $$x_n$$ is 3 less than the previous term, where $$n$$ and $$k$$ are positive integers and $$k<n$$. If $$x_0=x_n=0$$ and if $$x_k=15$$, what is the value of $$n$$?

A. 5
B. 6
C. 9
D. 10
E. 15

How can i approach these kind of problems???
[Reveal] Spoiler: OA

_________________

Last edited by Bunuel on 26 Jan 2012, 04:38, edited 2 times in total.
Edited the question
Math Expert
Joined: 02 Sep 2009
Posts: 38908
Followers: 7740

Kudos [?]: 106264 [15] , given: 11618

Show Tags

26 Jan 2012, 04:19
15
KUDOS
Expert's post
7
This post was
BOOKMARKED
kotela wrote:
In the sequence $$x_0, \ x_1, \ x_2, \ ... \ x_n$$, each term from $$x_1$$ to $$x_k$$ is 3 greater than the previous term, and each term from $$x_{k+1}$$ to $$x_n$$ is 3 less than the previous term, where $$n$$ and $$k$$ are positive integers and $$k<n$$. If $$x_0=x_n=0$$ and if $$x_k=15$$, what is the value of $$n$$?

A.5
B. 6
C. 9
D. 10
E. 15

How can i approach these kind of problems???

Probably the easiest way will be to write down all the terms in the sequence from $$x_0=0$$ to $$x_n=0$$. Note that each term from from $$x_0=0$$ to $$x_k=15$$ is 3 greater than the previous and each term from $$x_{k+1}$$ to $$x_n$$ is 3 less than the previous term:

So we'll have: $$x_0=0$$, 3, 6, 9, 12, $$x_k=15$$, 12, 9, 6, 3, $$x_n=0$$. So we have 11 terms from $$x_0$$ to $$x_n$$ thus $$n=10$$.

Hope it helps.
_________________
Director
Joined: 28 Jul 2011
Posts: 549
Location: United States
GPA: 3.86
WE: Accounting (Commercial Banking)
Followers: 3

Kudos [?]: 238 [0], given: 16

Re: In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk [#permalink]

Show Tags

26 Jan 2012, 04:37
Thanks Bunnel.....awesome explanation
_________________

Manager
Joined: 16 Feb 2012
Posts: 232
Concentration: Finance, Economics
Followers: 7

Kudos [?]: 327 [0], given: 121

Re: In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk [#permalink]

Show Tags

28 Apr 2012, 14:12
If we have 11 terms why is the answer 10? Which number do we don't count?
_________________

Kudos if you like the post!

Failing to plan is planning to fail.

Math Expert
Joined: 02 Sep 2009
Posts: 38908
Followers: 7740

Kudos [?]: 106264 [0], given: 11618

Re: In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk [#permalink]

Show Tags

28 Apr 2012, 15:59
Stiv wrote:
If we have 11 terms why is the answer 10? Which number do we don't count?

# of terms from $$x_0$$ to $$x_n$$ is indeed 11 but we are asked about the value of $$n$$, which is 10: $$x_0, \ x_1, \ x_2, \ ... \ x_{10}$$.

Hope it's clear.
_________________
Manager
Status: Private GMAT Tutor
Joined: 22 Oct 2012
Posts: 74
Location: India
Concentration: Economics, Finance
Schools: IIMA (A)
GMAT 1: 780 Q51 V47
Followers: 44

Kudos [?]: 174 [0], given: 36

Re: In the sequence X0, X1, X2 ......Xn each terms from X1 to Xk [#permalink]

Show Tags

25 Oct 2012, 18:40
1
This post was
BOOKMARKED
carcass wrote:
In the sequence $$X0$$, $$X1$$, $$X2$$ ......$$Xn$$ each terms from $$X1$$ to$$Xk$$ is 3 greater than the previous term, and each term from $$Xk+1$$ to $$Xn$$ is 3 less than the previous term, where$$N$$ and $$K$$ are positive integers and $$k < n$$. If $$X0$$ $$=$$ $$Xn = 0$$, what is the value of $$N$$ ?

(A) 5
(B) 6
(C) 9
(0) 10
(E) 15

This was tough, with a lot of information...........

If N=2k , all the conditions given in the question will be satisfied. Thus, any even natural number can be the answer. Of the options, both 6 and 10 can be the answers.

Cheers,
CJ
_________________
Intern
Joined: 03 Oct 2010
Posts: 4
Followers: 0

Kudos [?]: 0 [0], given: 2

Re: In the sequence X0, X1, X2 ......Xn each terms from X1 to Xk [#permalink]

Show Tags

26 Oct 2012, 02:13
The answer should be an odd positive integer. For example:
X0=0 and Xn=0 Then the series is: 0 3 6 3 0 (k=3 and n=5) or 0 3 6 9 6 3 0(k=4 and n=7).
So, either A or C or E could be the answer.
Correct me if I am missing something here.
Manager
Status: Private GMAT Tutor
Joined: 22 Oct 2012
Posts: 74
Location: India
Concentration: Economics, Finance
Schools: IIMA (A)
GMAT 1: 780 Q51 V47
Followers: 44

Kudos [?]: 174 [2] , given: 36

Re: In the sequence X0, X1, X2 ......Xn each terms from X1 to Xk [#permalink]

Show Tags

26 Oct 2012, 02:17
2
KUDOS
venuvm wrote:
The answer should be an odd positive integer. For example:
X0=0 and Xn=0 Then the series is: 0 3 6 3 0 (k=3 and n=5) or 0 3 6 9 6 3 0(k=4 and n=7).
So, either A or C or E could be the answer.
Correct me if I am missing something here.

Hi,

We are asked the value of N, not the number of elements in the series. Please note that if there are odd number of elements in the series, value of N will all always be even. This is because the first element is X0; which means number total number of elements in the series are N+1.

Cheers,
CJ
_________________
Senior Manager
Joined: 13 Aug 2012
Posts: 464
Concentration: Marketing, Finance
GPA: 3.23
Followers: 26

Kudos [?]: 467 [0], given: 11

Re: In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk [#permalink]

Show Tags

20 Dec 2012, 03:55
1
This post was
BOOKMARKED
x0 = 0
x1 = 3
x2 = 6
x3 = 9
x4 = 12
x5 = 15 Thus, k=5

The number of terms from 0 to k is equal the number of terms from n to k.

n-k = k-0
n-5 = 5
n = 10

_________________

Impossible is nothing to God.

Intern
Joined: 22 Jul 2010
Posts: 32
Followers: 0

Kudos [?]: 35 [0], given: 94

Show Tags

29 Aug 2013, 10:20
Bunuel wrote:
kotela wrote:
In the sequence $$x_0, \ x_1, \ x_2, \ ... \ x_n$$, each term from $$x_1$$ to $$x_k$$ is 3 greater than the previous term, and each term from $$x_{k+1}$$ to $$x_n$$ is 3 less than the previous term, where $$n$$ and $$k$$ are positive integers and $$k<n$$. If $$x_0=x_n=0$$ and if $$x_k=15$$, what is the value of $$n$$?

A.5
B. 6
C. 9
D. 10
E. 15

How can i approach these kind of problems???

Probably the easiest way will be to write down all the terms in the sequence from $$x_0=0$$ to $$x_n=0$$. Note that each term from from $$x_0=0$$ to $$x_k=15$$ is 3 greater than the previous and each term from $$x_{k+1}$$ to $$x_n$$ is 3 less than the previous term:

So we'll have: $$x_0=0$$, 3, 6, 9, 12, $$x_k=15$$, 12, 9, 6, 3, $$x_n=0$$. So we have 11 terms from $$x_0$$ to $$x_n$$ thus $$n=10$$.

Hope it helps.

Hi Bunuel

I felt this question was wrongly framed since it does not mention the relation between $$x_k$$ and $$x_{k+1}$$

There could be many series in that way and this question will have 3 answers

1)D-the same explanation you had given
2)E-the series would be 0 3 6 9 12 15=$$x_k$$ 27=$$x_{k+1}$$ 24 21 18 15 12 9 6 3 0
3)C-the series would be 0 3 6 9 12 15 9 6 3 0

Please correct me if I am wrong...
Math Expert
Joined: 02 Sep 2009
Posts: 38908
Followers: 7740

Kudos [?]: 106264 [0], given: 11618

Show Tags

29 Aug 2013, 10:25
Expert's post
1
This post was
BOOKMARKED
mitmat wrote:
Bunuel wrote:
kotela wrote:
In the sequence $$x_0, \ x_1, \ x_2, \ ... \ x_n$$, each term from $$x_1$$ to $$x_k$$ is 3 greater than the previous term, and each term from $$x_{k+1}$$ to $$x_n$$ is 3 less than the previous term, where $$n$$ and $$k$$ are positive integers and $$k<n$$. If $$x_0=x_n=0$$ and if $$x_k=15$$, what is the value of $$n$$?

A.5
B. 6
C. 9
D. 10
E. 15

How can i approach these kind of problems???

Probably the easiest way will be to write down all the terms in the sequence from $$x_0=0$$ to $$x_n=0$$. Note that each term from from $$x_0=0$$ to $$x_k=15$$ is 3 greater than the previous and each term from $$x_{k+1}$$ to $$x_n$$ is 3 less than the previous term:

So we'll have: $$x_0=0$$, 3, 6, 9, 12, $$x_k=15$$, 12, 9, 6, 3, $$x_n=0$$. So we have 11 terms from $$x_0$$ to $$x_n$$ thus $$n=10$$.

Hope it helps.

Hi Bunuel

I felt this question was wrongly framed since it does not mention the relation between $$x_k$$ and $$x_{k+1}$$

There could be many series in that way and this question will have 3 answers

1)D-the same explanation you had given
2)E-the series would be 0 3 6 9 12 15=$$x_k$$ 27=$$x_{k+1}$$ 24 21 18 15 12 9 6 3 0
3)C-the series would be 0 3 6 9 12 15 9 6 3 0

Please correct me if I am wrong...

Not so.

Stem says: each term from $$x_{k+1}$$ to $$x_n$$ is 3 less than the previous term. So, $$x_{k+1}$$ is 3 less than the previous term, which is $$x_k$$.

Hope it's clear.
_________________
Intern
Joined: 22 Jul 2010
Posts: 32
Followers: 0

Kudos [?]: 35 [0], given: 94

Show Tags

29 Aug 2013, 10:33
Hope it helps.[/quote]

Hi Bunuel

I felt this question was wrongly framed since it does not mention the relation between $$x_k$$ and $$x_{k+1}$$

There could be many series in that way and this question will have 3 answers

1)D-the same explanation you had given
2)E-the series would be 0 3 6 9 12 15=$$x_k$$ 27=$$x_{k+1}$$ 24 21 18 15 12 9 6 3 0
3)C-the series would be 0 3 6 9 12 15 9 6 3 0

Please correct me if I am wrong...[/quote]

Not so.

Stem says: each term from $$x_{k+1}$$ to $$x_n$$ is 3 less than the previous term. So, $$x_{k+1}$$ is 3 less than the previous term, which is $$x_k$$.

Hope it's clear.[/quote]

Ah that's true...thanks for the Clarification
Manager
Joined: 09 Nov 2012
Posts: 66
Followers: 0

Kudos [?]: 129 [0], given: 40

Re: In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk [#permalink]

Show Tags

18 Oct 2013, 01:51
I tried to use a standard linear sequence equation (Sn = k(n) + x, where 'k' is the constant difference between terms and x is another constant) to solve this:

x0, x1, x2, .......x(k), x(k+1),.....xn

For the first half of the sequence, the linear sequence would be Sn = k(n) + x => Sn =3(n) + x
Given: S0 or x0 = 0, therefore, 0 = 3(0) + x => x=0
Therefore Sn = 3(n).
Given: x15 = 15 = 3(n) => n = 5. So there are 5 terms in the first half of the sequence.

I couldn't set up the sequence for the 2nd half. Can someone help me?
Senior Manager
Joined: 08 Apr 2013
Posts: 276
Followers: 1

Kudos [?]: 27 [0], given: 27

Re: In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk [#permalink]

Show Tags

31 Oct 2013, 01:42
this is wron question which can not be from og. pls, show the question number in which og books.

what is relation between x(k+1) and x (k)

there is no relation and the question is wrong.
_________________

If anyone in this gmat forum is in England,Britain, pls, email to me, (thanghnvn@gmail.com) . I have some questions and need your advise. Thank a lot.

Math Expert
Joined: 02 Sep 2009
Posts: 38908
Followers: 7740

Kudos [?]: 106264 [0], given: 11618

Re: In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk [#permalink]

Show Tags

31 Oct 2013, 02:03
vietmoi999 wrote:
this is wron question which can not be from og. pls, show the question number in which og books.

what is relation between x(k+1) and x (k)

there is no relation and the question is wrong.

There is nothing wrong with the question. It's from The Official Guide for GMAT Quantitative Review, not sure about the question # though.

As for $$x_{k}$$ and $$x_{k+1}$$, they are consecutive terms in the sequence.
_________________
Senior Manager
Joined: 29 Oct 2013
Posts: 296
Concentration: Finance
GPA: 3.7
WE: Corporate Finance (Retail Banking)
Followers: 16

Kudos [?]: 415 [0], given: 197

Re: In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk [#permalink]

Show Tags

02 Apr 2014, 22:16
Bunuel wrote:
vietmoi999 wrote:
this is wron question which can not be from og. pls, show the question number in which og books.

what is relation between x(k+1) and x (k)

there is no relation and the question is wrong.

There is nothing wrong with the question. It's from The Official Guide for GMAT Quantitative Review, not sure about the question # though.

As for $$x_{k}$$ and $$x_{k+1}$$, they are consecutive terms in the sequence.

Yes, Bunuel is right it is from OG Quant Review 2nd Edition. The Questions No is 131 in problem solving section on page no. 78
_________________

My journey V46 and 750 -> http://gmatclub.com/forum/my-journey-to-46-on-verbal-750overall-171722.html#p1367876

Senior Manager
Joined: 29 Oct 2013
Posts: 296
Concentration: Finance
GPA: 3.7
WE: Corporate Finance (Retail Banking)
Followers: 16

Kudos [?]: 415 [2] , given: 197

Re: In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk [#permalink]

Show Tags

02 Apr 2014, 23:00
2
KUDOS
1
This post was
BOOKMARKED
Here is an algebraic approach in case the GMAT tests us with a similar question where the no of elements are too big to list out and solve.

Since each term in the sequence from X1 to Xk is 3 greater than the previous term, we can right a general formula,

Xk=X0+k*3, inserting the values Xk=15, X0=0
15=0+k*3
k=5

Now each term from Xk to Xn is 3 less than the previous term, the general formula will be

Xn=Xk-(n-k)*3, inserting the values Xn=0, Xk=15, k=5
0=15-(n-5)*3
0=15-3*n+15
3*n=30
n=10

Hope it makes sense
_________________

My journey V46 and 750 -> http://gmatclub.com/forum/my-journey-to-46-on-verbal-750overall-171722.html#p1367876

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 15474
Followers: 649

Kudos [?]: 209 [0], given: 0

Re: In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk [#permalink]

Show Tags

04 Apr 2015, 00:24
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.
_________________
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 15474
Followers: 649

Kudos [?]: 209 [0], given: 0

Re: In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk [#permalink]

Show Tags

05 May 2016, 01:50
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.
_________________
Senior Manager
Status: Professional GMAT Tutor
Affiliations: AB, cum laude, Harvard University (Class of '02)
Joined: 10 Jul 2015
Posts: 350
Location: United States (CA)
GMAT 1: 770 Q47 V48
GMAT 2: 730 Q44 V47
GRE 1: 337 Q168 V169
WE: Education (Education)
Followers: 59

Kudos [?]: 352 [1] , given: 45

Re: In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk [#permalink]

Show Tags

10 Aug 2016, 19:39
1
KUDOS
Top Contributor
Attached is a visual that should help.
Attachments

Screen Shot 2016-08-10 at 7.23.00 PM.png [ 89.74 KiB | Viewed 4145 times ]

_________________

Harvard grad and 770 GMAT scorer, offering high-quality private GMAT tutoring, both in-person and via Skype, since 2002.

McElroy Tutoring

Re: In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk   [#permalink] 10 Aug 2016, 19:39

Go to page    1   2    Next  [ 28 posts ]

Similar topics Replies Last post
Similar
Topics:
25 In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk 10 13 Nov 2016, 17:07
11 In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk 10 13 Apr 2016, 03:09
13 If the sequence x1, x2, x3, ... xn, ... is such that x1 = 3 4 21 Sep 2016, 18:37
22 If the sequence x_1, x_2, x_3, …, x_n, ... is such that x_1 9 01 Jan 2017, 14:03
27 If the sequence x1, x2, x3, …, xn, … is such that x1 = 3 13 24 May 2014, 06:57
Display posts from previous: Sort by