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

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Carcass

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In a certain sequence, each term beyond the second term is e [#permalink]

17 Sep 2017, 02:42

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This question is part of GREPrepClub - The Questions Vault Project

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.

$\frac{41}{2}$

Official Answer

D

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181charan

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

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.

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IlCreatore

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

29 Oct 2017, 01:42

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

28 Aug 2019, 05:49

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Carcass wrote:

This question is part of GREPrepClub - The Questions Vault Project

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.

$\frac{41}{2}$

Here,

$A1$ and

$A3$ are positive integer.

$A3 = \frac{{A1 + A2}}{2}$

For A3 to be positive integer, there can be 2 options

1. Both

$A1$ and

$A2$ are odd integer.

2. Both

$A1$ and

$A2$ are even integer.

and

$A3$ = always an even integer.

Let check for

$A4$ &

$A5$

$A4 = \frac{{A3 + A2}}{2}$

and

$A5 = \frac{{A4 + A3}}{2} = \frac{{3A3 + A2}}{4} ( since, A4 = \frac{{A3 + A2}}{2})$

Clearly, A5 cannot have a denominator greater than 4 ,

Hence option D

*** try putting values for A3 and A2 ( A3 =even integer and A2 =even or odd integer) , then A5 = all of the options except option D

aniruddha23

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In a certain sequence, each term beyond the second term is e [#permalink]

22 Jul 2021, 04:29

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Carcass wrote:

This question is part of GREPrepClub - The Questions Vault Project

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.

$\frac{41}{2}$

$A3 = \frac{A1+A2}{2}$

$A4 = \frac{A3+A2}{2}$

$A5 = \frac{A4+A3}{2}$

$A2 = 2A3 - A1$ (from 1st equation)

$A5 = \frac{1}{2}(\frac{A3+A2}{2}+\frac{A3}{2})$

$A5 = \frac{1}{2}(\frac{3}{2}A3+\frac{1}{2}A2)$

$A5 = \frac{1}{2}(\frac{3}{2}A3+\frac{1}{2}(2A3 - A1))$

Simplifying,

$A5 = \frac{1}{4}(5A3 - A1)$
Since, A1 and A3 are positive integers, the expression inside the bracket will always be an integer value. It can be equal to -9, 0, 9, 82 so that options A,B,D and E are possible.
Only option not possible is (D).

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In a certain sequence, each term beyond the second term is e [#permalink]

15 Aug 2021, 20:38

1

pranab223 wrote:

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.

$\frac{41}{2}$ [/quote]

Here,

$A1$ and

$A3$ are positive integer.

$A3 = \frac{{A1 + A2}}{2}$

For A3 to be positive integer, there can be 2 options

1. Both

$A1$ and

$A2$ are odd integer.

2. Both

$A1$ and

$A2$ are even integer.

and

$A3$ = always an even integer.

Let check for

$A4$ &

$A5$

$A4 = \frac{{A3 + A2}}{2}$

and

$A5 = \frac{{A4 + A3}}{2} = \frac{{3A3 + A2}}{4} ( since, A4 = \frac{{A3 + A2}}{2})$

Clearly, A5 cannot have a denominator greater than 4 ,

Hence option D

*** try putting values for A3 and A2 ( A3 =even integer and A2 =even or odd integer) , then A5 = all of the options except option D[/quote]

This problem is ill defined, because the question does not say if A2 is an integer or not. If A2 can be a fractional, then the conclusion in the above argument is wrong:
"Clearly, A5 cannot have a denominator greater than 4"

aniruddha23

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

15 Aug 2021, 23:03

2

Quote:

This problem is ill defined, because the question does not say if A2 is an integer or not. If A2 can be a fractional, then the conclusion in the above argument is wrong:
"Clearly, A5 cannot have a denominator greater than 4"

But the question mentions that A1 and A3 are positive integers.

$A3 = \frac{(A1 + A2)}{2}$
If A2 were to be fractional, A3 can't be an integer as it would go against what the question states.
So, A2 can't be fractional.

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

16 Jan 2025, 16:22

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

16 Jan 2025, 16:22

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