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

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26 Jan 2012, 03:15

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69% (02:36) correct
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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 \(n\)?

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 \(n=10\).

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

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25 Oct 2012, 17:40

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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.

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

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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.

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

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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.

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

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 \(x_k\).

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 \(x_k\).

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

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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?

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

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03 Apr 2015, 23:24

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

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05 May 2016, 00:50

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