It is currently 20 Mar 2018, 22:17

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

Intern
Joined: 18 Mar 2017
Posts: 42
Re: In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk [#permalink]

Show Tags

02 Apr 2017, 09:51
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.

Bunuel: if each term from $$x_{k+1}$$ to $$x_n$$ is 3 less than the previous term and if we have k=15, why don't we start at $$x_{15+1}$$ = $$x_{16}$$? This would then change the numbers in the sequence to $$x_k=16$$ = 13, 10, 7, 4, 1 ???

Intern
Joined: 18 Mar 2017
Posts: 42
Re: In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk [#permalink]

Show Tags

19 Apr 2017, 07:30
Does Bunuel or anyone else have a reply to my above named question?

Thank you very much!
gentler
Math Expert
Joined: 02 Sep 2009
Posts: 44352
Re: In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk [#permalink]

Show Tags

19 Apr 2017, 08:02
guenthermat 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.

Bunuel: if each term from $$x_{k+1}$$ to $$x_n$$ is 3 less than the previous term and if we have k=15, why don't we start at $$x_{15+1}$$ = $$x_{16}$$? This would then change the numbers in the sequence to $$x_k=16$$ = 13, 10, 7, 4, 1 ???

How did you get that k=15? We are given that $$x_k=15$$, not k = 15. Also, the fact that $$x_k=15$$ does not mean that $$x_{k+1}=16$$. We are told that each term from $$x_{k+1}$$ to $$x_n$$ is 3 less than the previous term. Thus $$x_{k+1}$$ is 3 less than the previous term which is $$x_k=15$$, so $$x_{k+1}=12$$
_________________
Intern
Joined: 18 Mar 2017
Posts: 42
In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk [#permalink]

Show Tags

21 Apr 2017, 11:35
Bunuel wrote:

How did you get that k=15? We are given that $$x_k=15$$, not k = 15. Also, the fact that $$x_k=15$$ does not mean that $$x_{k+1}=16$$. We are told that each term from $$x_{k+1}$$ to $$x_n$$ is 3 less than the previous term. Thus $$x_{k+1}$$ is 3 less than the previous term which is $$x_k=15$$, so $$x_{k+1}=12$$

But if we are told that (a) $$x_k=15$$ and that (b) each term from $$x_{k+1}$$ to $$x_n$$ is 3 less, then the term after $$x_k=15$$ – so $$x_{k+1}$$ – will be the starting term for this reduction of 3, won't it? But if you say that the sequence is 12, $$x_k=15$$, 12, then you already start reducing by 3 as of $$x_k=15$$ which is not $$x_{k+1}$$. Do you understand my issue?
Math Expert
Joined: 02 Sep 2009
Posts: 44352
Re: In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk [#permalink]

Show Tags

21 Apr 2017, 11:49
guenthermat wrote:
Bunuel wrote:

How did you get that k=15? We are given that $$x_k=15$$, not k = 15. Also, the fact that $$x_k=15$$ does not mean that $$x_{k+1}=16$$. We are told that each term from $$x_{k+1}$$ to $$x_n$$ is 3 less than the previous term. Thus $$x_{k+1}$$ is 3 less than the previous term which is $$x_k=15$$, so $$x_{k+1}=12$$

But if we are told that (a) $$x_k=15$$ and that (b) each term from $$x_{k+1}$$ to $$x_n$$ is 3 less, then the term after $$x_k=15$$ – so $$x_{k+1}$$ – will be the starting term for this reduction of 3, won't it? But if you say that the sequence is 12, $$x_k=15$$, 12, then you already start reducing by 3 as of $$x_k=15$$ which is not $$x_{k+1}$$. Do you understand my issue?

Each term from $$x_{k+1}$$ is 3 less than the previous term. So, $$x_{k+1}$$ is the first term which is 3 less than the previous term.
_________________
Intern
Joined: 18 Mar 2017
Posts: 42
Re: In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk [#permalink]

Show Tags

21 Apr 2017, 11:58
Now it makes perfect sense - thanks Bunuel!
Manager
Joined: 21 Jun 2017
Posts: 78
Re: In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk [#permalink]

Show Tags

07 Oct 2017, 08:40
mydreammba 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???

Working with the givens, X0 is your starting point. When you see this kind of sequence problem, it is best to just write the numbers out as if it were on a number line - helps with organization.

N is just a variable which represents the integers place in line. N is not related to the value of the sequencing digits. If it helps, personify math, and imagine these digits are waiting in line, and N is their ticket number.

X0, X1, X2, X3, X4, Xk (or X5), X6, X7, X8, X9, X10(Xn)
------------------------------------------
0, 3, 6 , 9, 12, 15 12 9 6 3 0

So the value of Xn is 0; but the question asks for the value of N alone, meaning its place in line.

Therefore, excluding X0, because 0 is not a value, we can conclude that the value for N is 10.

Re: In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk   [#permalink] 07 Oct 2017, 08:40

Go to page   Previous    1   2   [ 27 posts ]

Display posts from previous: Sort by