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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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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 [#permalink]
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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Re: In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk [#permalink]
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sri30kanth wrote:
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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Re: In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk [#permalink]
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 $$n=10$$.

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 ???

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Re: In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk [#permalink]
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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 $$n=10$$.

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 ???

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

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