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sasyaharry
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28 Oct 2017, 11:58
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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:
65%
(hard)
Question Stats:
38% (02:34) correct
63%
(02:08)
wrong
based on 8
sessions
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All the terms of a certain sequence \(x_1,x_2,........x_n\) are positive integers. The \(n^t^h\) term (n>1) of the sequence is given by the formula:
\(x_n =x_(n-1) + 1\) (If \(x_(n-1)\) is even)
\(x_n =x_(n-1) + 3\) (If \(x_(n-1)\) is odd)
What is the value of \(x_1 + x_6\)?
(1) The second term of the sequence is 3
(2) Two of the first three terms of the sequence are even and add up to 8
\(x_n =x_(n-1) + 1\) (If \(x_(n-1)\) is even)
\(x_n =x_(n-1) + 3\) (If \(x_(n-1)\) is odd)
What is the value of \(x_1 + x_6\)?
(1) The second term of the sequence is 3
(2) Two of the first three terms of the sequence are even and add up to 8
Archived Topic
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This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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28 Oct 2017, 12:09
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sasyaharry
Odd numbers are being added to \(x_{n-1}\), so if \(x_{n-1}\) is even then \(x_n\) is odd and if \(x_{n-1}\) is odd then \(x_n\) is even
to know the value of \(x_1+x_6\) we only need value of \(x_1\) remaining can be found out through the formula given in the question stem
Statement 1: implies \(x_2=x_n=3=odd\), so \(x_{n-1}=x_1=even\). we can substitute the value of \(x_2\) in the equation \(x_n=x_{n-1}+1\) to get \(x_1\). Sufficient
Statement 2: the sequence will have alternate odd and even numbers, so if out of 1st three two are even, this implies \(x_1\) is even, \(x_2\) is odd and \(x_3\) is even
given \(x_1+x_3=8 =>x_1+x_2+3=8\)
or \(x_1+x_1+1+3=8\), so \(x_1\) can be calculated. Sufficient
Option D
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Data Sufficiency (DS) Forum
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Thank you for understanding, and happy exploring!
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