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

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

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.

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

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.

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)

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 dave13

Writing down the terms helps(Here, \(x_k\) is nothing but \(x_5\))

\(x_0\)---\(x_1\)---\(x_2\)---\(x_3\)---\(x_4\)---\(x_5\)---\(x_6\)---\(x_7\)---\(x_8\)---\(x_9\)---\(x_{10}\)
0------3-----6-----9-----12----15----12-----9-----6-----3------0

Hope this helps you!

x0 = 0, 3, 6, 9, 12, 15, 12, 9, 6, 3, 0 = xn
now count you will get correct value of n

[quote="Bunuel"]The Official Guide For GMAT® Quantitative Review, 2ND Edition

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

If someone can help I hv a doubt. My approach was this:
Since x0 = 0 and each term from x1 to xk is 3 greater than the previous term, then xk = 0 + (k-1)(3). {Nth term = a+(n-1)d}
Since xk = 15, then 15 = 3k-3 and k = 6
. Since each term from xk + 1 to xn is 3 less than the previous term, then xn = xk − (n −1)(3). Substituting the known values for xk, xn,
gives 0 = 15 − (n − 1)(3), from which it follows that 0=15-3n+3 ====>3n=18 ====>n=6
Since N is 6 terms ahead of K ====> N=6+6=12 .

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

I have a question on the official answer given by GMAC:

Since x0 = 0 and each term from x1 to xk is 3 greater than the previous term, then xk = 0 + (k)(3). Since xk = 15, then 15 = 3k and k = 5. Since each term from xk + 1 to xn is 3 less than the previous term, then xn = xk− (n− k) (3). Substituting the known values for xk, xn, and k gives 0 = 15 − (n − 5) (3), from which it follows that 3n = 30 and n = 10.

In the bold highlighted expression, my question is how come it is (k)(3)?

As per the arithmetic progression formula an = a1 + (n-1)d the equation should have been xk = 0 + (k-1)(3) right?