In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk

Question

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

Bunuel · Accepted Answer

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.

PareshGmat · Answer

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

sri30kanth · Answer

Bunuel,

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

Bunuel · Answer

Yes, the number of terms is 11 but we start from  x_0 , not from  x_1 , hence n=10.

mcelroytutoring · Answer

Attached is a visual that should help.

guenthermat · Answer

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!

Bunuel · Answer

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 \(x_{k+1}=12\)

guenthermat · Answer

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?

Bunuel · Answer

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.

CyberStein · Answer

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)

CEdward · Answer

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

(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

EshaanTripathi · Answer

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

PGTLrowanhand · Answer

Here's my video solution:  https://www.youtube.com/watch?v=1aRw_KBtXkc

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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