Bunuel wrote:
All the terms of a certain sequence \(x_1, \ x_2, \ ..., \ x_n\) are positive integers. The \(n^{th}\) 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
Given- \(x_n =x_{(n-1)} + 1\) (If \(x_{(n-1)}\) is even)
\(x_n =x_{(n-1)} + 3\) (If \(x_{(n-1)}\) is odd) - \(x_1, \ x_2, \ ..., \ x_n\) are positive integers
Statement 1The second term of the sequence is 3\(x_2\) = 3
Now we do not know if \(x_{n-1}\) is even or \(x_{n-1}\) is odd. Let's consider both the cases
Case 1: \(x_{n-1}\) is even
\(x_2 = x_1 + 1\)
\(x_1 +1 = 3\)
\(x_1 = 2\)
Case 2: \(x_{n-1}\) is odd
\(x_2 = x_1 + 3\)
\(x_1 +3 = 3\)
\(x_1 = 0\)
Case 2 is not possible as it is given that \(x_1\) is a positive integer. Hence \(x_1 = 2\)
Once we know \(x_1\), we can find all the terms in the sequence. Note, we don't have to find the terms as this is a DS question. However, we do have sufficient information if need be.
Hence this statement is sufficient. We can eliminate B, C and E.
Statement 2Two of the first three terms of the sequence are even and add up to 8
The first three terms of the sequence are \(x_1, x_2\) and \(x_3\)
Before we proceed analyzing the statement, let's have a quick look at the formula and see if we can infer anything
\(x_n =x_{(n-1)} + 1\) (If \(x_{(n-1)}\) is even)
Inference: If \(x_{(n-1)}\) is even, \(x_n\) = even + odd = odd\(x_n =x_{(n-1)} + 3\) (If \(x_{(n-1)}\) is odd)
Inference: If \(x_{(n-1)}\) is odd , \(x_n\) = odd + odd = evenCase 1:
\(x_1\) = odd | \(x_1\)
\(x_2\) = even | \(x_2\) = \(x_1\) + 3
\(x_3\) = odd | \(x_3\) = \(x_2\) + 1= \(x_1\) + 4
The only two terms that can add to even is \(x_1\) + \(x_3\)
\(x_1\) + \(x_1\) + 4 = 8
2\(x_1\) = 4
\(x_1\) = \(\frac{4}{2 } = 2\)
We have to discard this case, as \(x_1\) was assumed to be odd.
Case 2:
\(x_1\) = even| \(x_1\)
\(x_2\) = odd | \(x_2\) = \(x_1\) + 1
\(x_3\) = even| \(x_3\) = \(x_2\) + 3= \(x_1\) + 4
The only two terms that can add to even is \(x_1\) + \(x_3\)
\(x_1\) + \(x_1\) + 4 = 8
2\(x_1\) = 4
\(x_1\) = 2
Once we know \(x_1\), we can find all the terms in the sequence.
Hence this statement is sufficient.
Option D