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

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

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14 Dec 2010, 17:40
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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

OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/in-the-seque ... 68359.html
[Reveal] Spoiler: OA
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Re: In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk [#permalink]

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14 Dec 2010, 18:00
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Expert's post
tonebeeze wrote:
Hello all,

I am hoping you all can provide me with some help with Quant Review PS #131. The explanation in the book was not clear enough for me. Hopefully you all can help break this problem down easily, as I am looking for better approaches to sequence problem types. Thank you.

131. 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

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 $$n=10$$.

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

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14 Dec 2010, 18:05
Perfect explanation Bunnuel. Much more clearer. My issue was that I did not fully comprehend exactly what the question was asking. By listing everything out, the problem is actually very simple. Thank you!
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Re: In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk [#permalink]

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14 Dec 2010, 21:18
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0,X1,X2,X3,X4,...15,

0,3,6,9,12,15,12,9,6,3,0

n = 10
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Re: In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk [#permalink]

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15 Dec 2010, 07:58
I alos just wrote down the values in the sequence. It took very little time and being able to see the numbers made it very easy to answer
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Re: In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk [#permalink]

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28 Feb 2011, 08:24
x0=0
X1=3
x2=6
.
.
x5=15
x6=12
x7=9
x8=6
x9=3
x10=0

Clearly n=10
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Re: In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk [#permalink]

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Updated on: 26 Jan 2012, 04:38
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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???
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Originally posted by mydreammba on 26 Jan 2012, 04:15.
Last edited by Bunuel on 26 Jan 2012, 04:38, edited 2 times in total.
Edited the question
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Joined: 02 Sep 2009
Posts: 44600
Re: In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk [#permalink]

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26 Jan 2012, 04:19
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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.
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Re: In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk [#permalink]

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28 Apr 2012, 14:12
If we have 11 terms why is the answer 10? Which number do we don't count?
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Re: In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk [#permalink]

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

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25 Oct 2012, 18:40
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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
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Re: In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk [#permalink]

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

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26 Oct 2012, 02:17
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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
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Re: In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk [#permalink]

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20 Dec 2012, 03:55
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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

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

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29 May 2013, 20:42
Following up on kind of an old post, but I thought I'd try just in case. I'm confused as to why the answer is ten and NOT eleven. Thanks.
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Re: In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk [#permalink]

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30 May 2013, 01:01
xtineb wrote:
Following up on kind of an old post, but I thought I'd try just in case. I'm confused as to why the answer is ten and NOT eleven. Thanks.

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

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15 Jun 2013, 07:26
I thought x(k+1) = 15+3 since it wasnt said that from x( k+1) to xn inclusive (!) is 3 less
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Re: In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk [#permalink]

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15 Jun 2013, 08:56
LalaB wrote:
I thought x(k+1) = 15+3 since it wasnt said that from x( k+1) to xn inclusive (!) is 3 less

Actually,it is mentioned :

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.....

It clearly means that starting from term $$x_k_+_1$$ till $$x_n$$, every term is 3 less than the previous one.
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Re: In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk [#permalink]

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

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29 Aug 2013, 10:25
Expert's post
1
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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.
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Re: In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk   [#permalink] 29 Aug 2013, 10:25

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