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In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk
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05 Mar 2014, 01:20
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The Official Guide For GMAT® Quantitative Review, 2ND EditionIn 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 Problem Solving Question: 131 Category: Algebra Sequences Page: 78 Difficulty: 600 GMAT Club is introducing a new project: The Official Guide For GMAT® Quantitative Review, 2ND Edition  Quantitative Questions ProjectEach week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution. We'll be glad if you participate in development of this project: 1. Please provide your solutions to the questions; 2. Please vote for the best solutions by pressing Kudos button; 3. Please vote for the questions themselves by pressing Kudos button; 4. Please share your views on difficulty level of the questions, so that we have most precise evaluation. Thank you!
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Re: In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk
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SOLUTIONIn 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 \(n=10\). Answer: D.
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Re: In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk
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05 Mar 2014, 19:25
Answer = D = 10 Series will be formed as below starting from x0 = 0 & ending up with x10 = 0 xk = 15 0, 3, 6, 9, 12, 15, 12, 9, 6, 3, 0
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Re: In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk
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26 Jul 2014, 21:17
Bunuel,
If the total terms ins sequence are 11, how is the no of terms n 10? Can you please explain? Thanks in advance.



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In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk
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Re: In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk
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27 Jul 2014, 05:53
Yes so the correct ans should be 11 which is not present in any of the options.



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Re: 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
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10 Aug 2016, 18:39
Attached is a visual that should help.
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Re: In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk
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02 Apr 2017, 08:51
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 \(n=10\). Answer: D. 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 ??? Thanks for your help!



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Re: In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk
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19 Apr 2017, 07:02
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 \(n=10\). Answer: D. 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 ??? Thanks for your help! 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 \(x_{k+1}=12\)
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In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk
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21 Apr 2017, 10:35
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?



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Re: In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk
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21 Apr 2017, 10:49
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.
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Re: In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk
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07 Oct 2017, 07:40
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.
Answer is (D)



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Re: In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk
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27 Mar 2018, 11:49
Bunuel wrote: SOLUTION
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 \(n=10\).
Answer: D. if \(x_n=0\) and we are asked to find the value of the value of \(n\) ? and there are total 11 terms, and \(x_n=0\) is last term .... why are we saying that \(x_n = x_10\) are we asked to find the last value of n why are we counting 10 terms only , ok we count from zero ? so ? if i count from zero there are 11 terms ... whats point to think is it 10 ? and the tenth term is 3 by the way right ? totally confused by the question so many questions becaus of only one question.... i got into the habit of mutliplying questions



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Re: In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk
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27 Mar 2018, 12:55
dave13 wrote: Bunuel wrote: SOLUTION
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 \(n=10\).
Answer: D. if \(x_n=0\) and we are asked to find the value of the value of \(n\) ? and there are total 11 terms, and \(x_n=0\) is last term .... why are we saying that \(x_n = x_10\) are we asked to find the last value of n why are we counting 10 terms only , ok we count from zero ? so ? if i count from zero there are 11 terms ... whats point to think is it 10 ? and the tenth term is 3 by the way right ? totally confused by the question so many questions becaus of only one question.... i got into the habit of mutliplying questions Hi dave13Writing down the terms helps(Here, \(x_k\) is nothing but \(x_5\)) \(x_0\)\(x_1\)\(x_2\)\(x_3\)\(x_4\)\(x_5\)\(x_6\)\(x_7\)\(x_8\)\(x_9\)\(x_{10}\) 03691215129630 Hope this helps you!
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