Series A(n) is such that i*A(i) = j*A(j) for any pair of

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Series A(n) is such that i*A(i) = j*A(j) for any pair of [#permalink]

07 Feb 2012, 03:52

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Series A(n) is such that i*A(i) = j*A(j) for any pair of positive integers (i, j). If A(1) is a positive integer, which of the following is possible?

I. 2*A(100) = A(99) + A(98)
II. A(1) is the only integer in the series
III. The series does not contain negative numbers

A) I only
B) II only
C) I & III only
D) II & III only
E) I, II & III

Will appreciate if anyone explains this question with an easy method.

Thanks & Regards
Vinni

[Reveal] Spoiler: OA

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Re: Series A(n) is such that i*A(i) = j*A(j) [#permalink]

07 Feb 2012, 04:02

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vinnik wrote:

Probably it should be sequence instead of series.

A sequence of numbers a_1, a_2, a_3, ... have the following properties: i*a_i=j*a_j and a_1=positive \ integer, so 1*a_1=2*a_2=3*a_3=4*a_4=5*a_5=...=positive \ integer.

We should determine whether the options given below can occur (notice that the question is which of the following COULD be true, not MUS be true).

I. 2a_{100}=a_{99}+a_{98} --> as 100a_{100}=99a_{99}=98a_{98}, then 2a_{100}=\frac{100}{99}a_{100}+\frac{100}{98}a_{100} --> reduce by a_{100} --> 2=\frac{100}{99}+\frac{100}{98} which is not true. Hence this option cannot be true.

II. a_1 is the only integer in the series. If a_1=1, then all other terms will be non-integers --> a_1=1=2a_2=3a_3=... --> a_2=\frac{1}{2}, a_3=\frac{1}{3}, a_4=\frac{1}{4}, and so on. Hence this option can be true.

III. The series does not contain negative numbers --> as given that a_1=positive \ integer=n*a_n, then a_n=\frac{positive \ integer}{n}=positive \ number, hence this option is always true.

Answer: D (II and III only).

Hope it's clear.
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Re: Series A(n) is such that i*A(i) = j*A(j) [#permalink]

07 Feb 2012, 04:34

Bunuel wrote:

A sequence of numbers a_1, a_2, a_3, ... have the following properties: i*a_i=j*a_j and a_1=positive \ integer, so 1*a_1=2*a_2=3*a_3=4*a_4=5*a_5=...=positive \ integer.

Bunuel,

I didnt get this part. I seem to misunderstood the q.stem

could u please clarify it?
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Re: Series A(n) is such that i*A(i) = j*A(j) [#permalink]

07 Feb 2012, 04:39

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LalaB wrote:

Bunuel wrote:

A sequence of numbers a_1, a_2, a_3, ... have the following properties: i*a_i=j*a_j and a_1=positive \ integer, so 1*a_1=2*a_2=3*a_3=4*a_4=5*a_5=...=positive \ integer.

Bunuel,

I didnt get this part. I seem to misunderstood the q.stem

could u please clarify it?

Sure.

Given: a_1=positive \ integer. Next, i*a_i=j*a_j, notice that we have the same multiple and the same index of a on both sides: 1*a_1=2*a_2, 2*a_2=3*a_3, a_3=4*a_4.... Hence, 1*a_1=2*a_2=3*a_3=4*a_4=5*a_5=...=positive \ integer (it equal to an integer since a_1=positive \ integer).

Hope it's clear.
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Re: Series A(n) is such that i*A(i) = j*A(j) for any pair of [#permalink]

07 Feb 2012, 09:41

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vinnik wrote:

First thing I want to understand is this relation: i*A(i) = j*A(j) for any pair of positive integers. I will take examples to understand it.

When i = 1 and j = 2, A(1) = 2*A(2)
So A(2) = A(1)/2

When i = 1 and j = 3, A(1) = 3*A(3)
So A(3) = A(1)/3

I see it now. The series is: A(1), A(1)/2, A(1)/3, A(1)/4 and so on...

II and III are easily possible. We can see that without any calculations.

II. A(1) is the only integer in the series
If A(1) = 1, then series becomes 1, 1/2, 1/3, 1/4 ... all fractions except A(1)

III. The series does not contain negative numbers
Again, same series as above applies. In fact, since A(1) is a positive integer, this must be true.

I. 2*A(100) = A(99) + A(98)
2*A(1)/100 = A(1)/99 + A(1)/98 (cancel A(1) from both sides)
2/100 = 1/99 + 1/98
Not true hence this is not possible

Answer (D)
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Re: Series A(n) is such that i*A(i) = j*A(j) for any pair of [#permalink]

07 Feb 2012, 11:59

got it at last:) thnx
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Re: Series A(n) is such that i*A(i) = j*A(j) for any pair of [#permalink]

08 Feb 2012, 08:28

Thanks Bunuel and Karishma for clearing my doubt. :-D

Regards
Vinni

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Re: Series A(n) is such that i*A(i) = j*A(j) for any pair of [#permalink]

21 May 2012, 08:21

Bunuel
could you please go thru this part one more time?
Cant get it

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Re: Series A(n) is such that i*A(i) = j*A(j) for any pair of [#permalink]

21 May 2012, 09:15

Galiya wrote:

Bunuel
could you please go thru this part one more time?
Cant get it

From 100a_{100}=99a_{99} --> a_{99}=\frac{100}{99}a_{100};

From 100a_{100}=98a_{98} --> a_{98}=\frac{100}{98}a_{100};

So, option I. 2a_{100}=a_{99}+a_{98} becomes: 2a_{100}=\frac{100}{99}a_{100}+\frac{100}{98}a_{100}.

Hope it's clear.
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Re: Series A(n) is such that i*A(i) = j*A(j) for any pair of [#permalink] 21 May 2012, 09:15

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