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

Question

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

A.  -\frac{9}{4}

B. 0

C.  \frac{9}{4}

D.  \frac{75}{8}

E.  \frac{41}{2}

Shobhit7 · Accepted Answer

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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GMATinsight · Answer

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

Shef08 · Answer

Hi I need help on this one. Can any of the experts look in to this..

Posted from my mobile device

DarkHorse2019 · Answer

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.

KarishmaB · Answer

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 =  /2 = (3*A3 + A2)/4

Since A2 and A3 are integers, you cannot have 8 in the denominator. 
Answer (D)

chacinluis · Answer

The issue is not getting it....it's getting it in 1:45 secs

HoudaSR · Answer

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

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