SandhyAvinash wrote:

All the terms of a certain sequence x1,x2,........xn are positive integers. The nth term (n>1) of the sequence is given by the formula:

xn =xn-1 + 1 (If xn-1 is even)

xn =xn-1 + 3 (If xn-1 is odd)

What is the value of x1 + x6?

(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

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 \(x_1\).

SufficientStatement 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.

SufficientOption

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