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In a certain sequence, each term beyond the second term is e

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In a certain sequence, each term beyond the second term is e [#permalink] New post 17 Sep 2017, 02:42
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In a certain sequence, each term beyond the second term is equal to the average of the previous two terms. If A1 and A3 are positive integers, which of the following is not a possible value of A5?


A. \frac{-9}{4}

B. zero

C. \frac{9}{4}

D. \frac{75}{8}

E. \frac{41}{2}
[Reveal] Spoiler: OA

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Re: In a certain sequence, each term beyond the second term is e [#permalink] New post 29 Oct 2017, 01:42
Any help here?
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Re: In a certain sequence, each term beyond the second term is e [#permalink] New post 11 Nov 2017, 14:53
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IlCreatore wrote:
Any help here?


a1, a2, a3 = (a1+a2)/2, a4 = (a3+a2)/2 = (3a2 + a1)/4 , a5 = (a3 + a4)/2 = (5a2 + 3a1)/8


a5 = (5a2 + 3a1) / 8.

Given to us that a1 is an integer and a3 is an integer.

if a3 is an integer that means sum of a1 and a2 is an even number, So say (a1 + a2) = 2x

hence a5 = ( 3(a2+a1) + 2(a2) ) / 8

a5 = ( (3*2x) + 2(a2) ) / 8

This shows us that the numerator is always an even number as every digit is a multiple of 2, hence sum of the two digits is an even number.

(even number)/8
hence if the denominator is 8 the numerator should be an even number. So the only number with an odd numerator and 8 as denominator is 75/8 hence it is a value not possible.
Re: In a certain sequence, each term beyond the second term is e   [#permalink] 11 Nov 2017, 14:53
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In a certain sequence, each term beyond the second term is e

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