Sequence

greenwich wrote:In the sequence X0, 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=0 and if Xk=15, what is the value of n?

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

What a question

If xk = 15 and x0 = 0, then the series must begin as under

0, 3, 6, 9, 12, 15, …

*In this continuity, x(k + 1) will be 12, and from here onwards, it would meet its end at 0 as under

… 12, 9, 6, 3, 0.

Hence, the series as a whole is a peculiar example of number symmetry

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

It has got 11 terms to it, hence n = [spoiler]10.


D[/spoiler]

Received the following PM from pallasy to respond:

pallasy wrote:

I was reading your response to this post (https://www.beatthegmat.com/sequence-t67057.html) and was wondering why you said "It has got 11 terms to it, hence [spoiler]n = 10[/spoiler]." What's the logic you used to arrive at that conclusion?

Response:

As the sequence X0, X1, X2, ..., Xn has got its first term represented by x0, second term represented by x1, with xn representing its 11th term, hence 11 must be equal to [spoiler]n + 1, or n = 10[/spoiler].