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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 [#permalink]

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New post 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 [#permalink]

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New post 19 Apr 2017, 06:30
Does Bunuel or anyone else have a reply to my above named question?

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

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New post 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 [#permalink]

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New post 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 [#permalink]

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New post 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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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

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New post 21 Apr 2017, 10:58
Now it makes perfect sense - thanks Bunuel!

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

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New post 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   [#permalink] 07 Oct 2017, 07:40

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