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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 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
[Reveal] Spoiler: OA

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

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New post 07 Feb 2012, 04:02
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vinnik wrote:
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


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) for any pair of [#permalink]

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New post 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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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]

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vinnik wrote:
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


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]

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New post 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]

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New post 08 Feb 2012, 08:28
Thanks Bunuel and Karishma for clearing my doubt. :-D

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

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New post 21 May 2012, 08:21
Bunuel
could you please go thru this part one more time?
Cant get it
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New post 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]

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New post 28 Nov 2015, 00:22
Bunuel wrote:

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.



I don't understand this part.
How could I know that A(1) = 1
The question stem mentioned only that "A(1) is a positive integer"

If A(1) = 2, then ---> 1*A(1) = 2*A(2) ---> A(2) = 1
Then II cannot be true.

Please tell me if I get something wrong.

Thanks

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New post 28 Nov 2015, 08:29
pakasaip wrote:
Bunuel wrote:

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.



I don't understand this part.
How could I know that A(1) = 1
The question stem mentioned only that "A(1) is a positive integer"

If A(1) = 2, then ---> 1*A(1) = 2*A(2) ---> A(2) = 1
Then II cannot be true.

Please tell me if I get something wrong.

Thanks


Please notice that it says "IF \(a_1=1\), ..." and also that the question asks which of the following is possible, so which of the following could be true.
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Re: Series A(n) is such that i*A(i) = j*A(j) for any pair of [#permalink]

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New post 30 Jun 2016, 03:37
vinnik wrote:
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


Right..only looks complicated(that's how you should think first for some confidence :lol: )
Let's give this a shot..

From the given equation we know that..
\(1*A(1) = k*A(k)\) ..where k is any positive integer..

Coming over to the premises..we'll deal with I at the end

II Possible..what if A(1)=1? every other term will be a fraction..so YES

III Always true..no explanation needed

I
because it's a "could be true" question..we won't give A(1) a value for this statement..and go with A(1) as..some number/fraction A(1)
\(2*A(100) = A(99) + A(98)\)

\(A(100) + A(100) = A(99) + A(98)\)
We know that..
\(1*A(1) = 100*A(100)\)
\(1*A(1) = 99*A(99)\)
and
\(1*A(1) = 98*A(98)\)

Using the expressions and transforming the main equation..

\(2*\frac{A(1)}{100} = \frac{A(1)}{99} + \frac{A(1)}{98}\)

\(\frac{A(1)}{50} = \frac{A(1)}{99} + \frac{A(1)}{98}\)

And we know that R.H.S. has no "5" in it...so this will NEVER be true.. :)
Answer (D)

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