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Math Expert V
Joined: 02 Sep 2009
Posts: 57082
In a certain sequence, each term beyond the second term is equal to th  [#permalink]

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15 00:00

Difficulty:   95% (hard)

Question Stats: 21% (02:53) correct 79% (02:31) wrong based on 116 sessions

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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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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]

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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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Re: In a certain sequence, each term beyond the second term is equal to th  [#permalink]

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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]

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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]

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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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