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In a certain sequence, each term beyond the second term is equal to th

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Bunuel
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In a certain sequence, each term beyond the second term is equal to th [#permalink] New post   15 May 2018, 02:47
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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 \(A_1\) and \(A_3\) are positive integers, which of the following is not a possible value of \(A_5\)?


A. \(-\frac{9}{4}\)

B. 0

C. \(\frac{9}{4}\)

D. \(\frac{75}{8}\)

E. \(\frac{41}{2}\)
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D
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Shobhit7
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Re: In a certain sequence, each term beyond the second term is equal to th [#permalink] New post   15 May 2018, 12:47
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Let the terms be
a1, a2, a3, a4 and a5

ATS,
a3=(a1+a2)/2 is a positive integer. Therefore, a1+a2 is even. Hence, both a1 and a2 are either even or odd.

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

As a1 & a2 are both either odd or even, Numerator i.e.(5*a2+3*a1) must be even.

Plugin Ans choices:
A. -9/4
So, (5*a2+3*a1)=-18, possible

B. 0
So, (5*a2+3*a1)=0, possible

C. 9/4
So, (5*a2+3*a1)=18, possible

D. 75/8
So, (5*a2+3*a1)=75, not possible

E. 41/2
So, (5*a2+3*a1)=82, possible

Hence, Ans D

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In a certain sequence, each term beyond the second term is equal to th [#permalink] New post   15 May 2018, 03:24
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Bunuel wrote:
In a certain sequence, each term beyond the second term is equal to the average of the previous two terms. If \(A_1\) and \(A_3\) are positive integers, which of the following is not a possible value of \(A_5\)?


A. \(-\frac{9}{4}\)

B. 0

C. \(\frac{9}{4}\)

D. \(\frac{75}{8}\)

E. \(\frac{41}{2}\)


Since \(A_1\) and \(A_3\) are both Integers so \(A_2\) must be integer as well

at the same time

\(A_1\) and \(A_2\) should both be either even or both should be Odd

\(A_1\) = Even
\(A_2\) = Even
\(A_3\) = Even/Odd
\(A_4\) = Even/ Odd / Integer2
\(A_5\) = Even / Odd/ Inteegr/4

i.e. denominator part of \(A_5\) can never be more than 4 hence

Answer: Option D
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Re: In a certain sequence, each term beyond the second term is equal to th [#permalink] New post   18 Jul 2019, 10:49
Hi I need help on this one. Can any of the experts look in to this..

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In a certain sequence, each term beyond the second term is equal to th [#permalink] New post   28 Jul 2019, 09:51
Shef08 wrote:
Hi I need help on this one. Can any of the experts look in to this..

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Dear Moderators chetan2u VeritasKarishma Bunuel Gladiator59 generis

Please can you explain this one?

I did not understand why A1 & A2 have to be both odd or both even.
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Re: In a certain sequence, each term beyond the second term is equal to th [#permalink] New post   30 Jul 2019, 04:41
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DarkHorse2019 wrote:
Shef08 wrote:
Hi I need help on this one. Can any of the experts look in to this..

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Dear Moderators chetan2u VeritasKarishma Bunuel Gladiator59 generis

Please can you explain this one?

I did not understand why A1 & A2 have to be both odd or both even.


The question tells you that A3 is a positive integer.

A3 = (A1 + A2)/2 = Integer

This means (A1 + A2) = 2*Integer = Even Integer

When will sum of two numbers be even? When both are even or both are odd.
Even + Even = Even
Odd + Odd = Even

Even + Odd = Odd

So A1 and A2 are both either even or both odd. Anyway, it doesn't matter.

You know that A1, A2 and A3 all are integers. A4 may not be an integer.
A4 = (A2 + A3)/2
A5 = (A3 + A4)/2 = [A3 + (A2 + A3)/2]/2 = (3*A3 + A2)/4

Since A2 and A3 are integers, you cannot have 8 in the denominator.
Answer (D)
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Re: In a certain sequence, each term beyond the second term is equal to th [#permalink] New post   14 Oct 2020, 21:21
The issue is not getting it....it's getting it in 1:45 secs
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Re: In a certain sequence, each term beyond the second term is equal to th [#permalink] New post   16 Apr 2023, 15:09
Bunuel wrote:
In a certain sequence, each term beyond the second term is equal to the average of the previous two terms. If \(A_1\) and \(A_3\) are positive integers, which of the following is not a possible value of \(A_5\)?


A. \(-\frac{9}{4}\)

B. 0

C. \(\frac{9}{4}\)

D. \(\frac{75}{8}\)

E. \(\frac{41}{2}\)



A3= (A1+A2)/2 -->A2= 2A3-A1 and since A1 and A3 are integers, A2 must also be an integer.

Also A4= (A2+A3)/2
And A5= (A3+A4)/2
Replacing A4 by (A2+A3)/2 in the second equation, we get A5= (3A3+A2)/4.

Since A3 and A2 are integers, we have A5= Some integer/4

Therefore A5 can be any integer divided by 4 or if the fraction is reduced it could be some integer divider by 2, or it could be 0 if the numerator is 0 which is an integer but it can't be an integer divided by 8.

The correct answer is D.
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Re: In a certain sequence, each term beyond the second term is equal to th [#permalink]
16 Apr 2023, 15:09
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