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

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In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk [#permalink] New post 26 Jan 2012, 03:15
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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???
[Reveal] Spoiler: OA

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Last edited by Bunuel on 26 Jan 2012, 03:38, edited 2 times in total.
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Re: Sequences [#permalink] New post 26 Jan 2012, 03:19
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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 \(n=10\).

Answer: D.

Hope it helps.
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Re: In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk [#permalink] New post 26 Jan 2012, 03:37
Thanks Bunnel.....awesome explanation
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Re: In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk [#permalink] New post 28 Apr 2012, 13:12
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] New post 28 Apr 2012, 14:59
Expert's post
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 terms from X1 to Xk [#permalink] New post 25 Oct 2012, 17:40
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.

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Re: In the sequence X0, X1, X2 ......Xn each terms from X1 to Xk [#permalink] New post 26 Oct 2012, 01:13
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 terms from X1 to Xk [#permalink] New post 26 Oct 2012, 01:17
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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.

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Re: In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk [#permalink] New post 20 Dec 2012, 02:55
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

Answer: 10
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Re: Sequences [#permalink] New post 29 Aug 2013, 09:20
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 \(n=10\).

Answer: D.

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: Sequences [#permalink] New post 29 Aug 2013, 09:25
Expert's post
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 \(n=10\).

Answer: D.

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: Sequences [#permalink] New post 29 Aug 2013, 09:33
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 :)
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Re: In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk [#permalink] New post 18 Oct 2013, 00:51
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?
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Re: In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk [#permalink] New post 31 Oct 2013, 00:42
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.
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Re: In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk [#permalink] New post 31 Oct 2013, 01:03
Expert's post
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.
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Re: In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk [#permalink] New post 02 Apr 2014, 21:16
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
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Re: In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk [#permalink] New post 02 Apr 2014, 22:00
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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
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Re: In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk [#permalink] New post 03 Apr 2015, 23:24
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Re: In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk   [#permalink] 03 Apr 2015, 23:24
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