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555-605 Level|   Sequences|                     
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Bunuel
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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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Bunuel,

If the total terms ins sequence are 11, how is the no of terms n 10? Can you please explain? Thanks in advance.

Yes, the number of terms is 11 but we start from \(x_0\), not from \(x_1\), hence n=10.
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Attached is a visual that should help.
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Screen Shot 2016-08-10 at 7.23.00 PM.png
Screen Shot 2016-08-10 at 7.23.00 PM.png [ 89.74 KiB | Viewed 71214 times ]

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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???

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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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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Bunuel


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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Bunuel


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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mydreammba
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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In the sequence x0, x1, x2, ... xnx0, x1, x2, ... xn, each term 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=0x0=xn=0 and if xk=15, what is the value of n?

(A) 5
(B) 6
(C) 9
(D) 10
(E) 15


x0 = 0, x1, x2,x3...xk, xk+1, ...xn = 0

So...

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

n = 10

D
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How I solved this in 30 secs

If Each term upto Xk is 3 greater than previous and Xk is 15 then total tems until xk should be 15/3 = 5
since its is coming back to zero with the the same difference to Xn, the other half of terms is also 5 hence, 5+5 = 10
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