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All the terms of a certain sequence x1,x2,........xn are positive...

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All the terms of a certain sequence x1,x2,........xn are positive... [#permalink]

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New post 28 Oct 2017, 11:58
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  75% (hard)

Question Stats:

36% (01:11) correct 64% (01:58) wrong based on 11 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
[Reveal] Spoiler: OA

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All the terms of a certain sequence x1,x2,........xn are positive... [#permalink]

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New post 28 Oct 2017, 12:09
sasyaharry 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 \(x_1\). 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

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All the terms of a certain sequence x1,x2,........xn are positive...   [#permalink] 28 Oct 2017, 12:09
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