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05 Mar 2014, 02:20
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
75% (02:11) correct
25%
(02:27)
wrong
based on 4150
sessions
History
Date
Time
Result
Not Attempted Yet
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
(A) 5
(B) 6
(C) 9
(D) 10
(E) 15
05 Mar 2014, 02:20
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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.
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.
General Discussion
05 Mar 2014, 20:25
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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
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
