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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 [#permalink]

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07 Feb 2012, 02:52
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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
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Re: Series A(n) is such that i*A(i) = j*A(j) [#permalink]

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07 Feb 2012, 03:02
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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) [#permalink]

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07 Feb 2012, 03:34
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]

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07 Feb 2012, 03:39
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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]

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07 Feb 2012, 08:41
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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

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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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07 Feb 2012, 10:59
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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08 Feb 2012, 07:28
Thanks Bunuel and Karishma for clearing my doubt.

Regards
Vinni
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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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21 May 2012, 07:21
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]

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21 May 2012, 08:15
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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27 Nov 2015, 23:22
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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Re: Series A(n) is such that i*A(i) = j*A(j) for any pair of [#permalink]

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28 Nov 2015, 07:29
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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30 Jun 2016, 02:37
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 )
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..

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Re: Series A(n) is such that i*A(i) = j*A(j) for any pair of   [#permalink] 30 Jun 2016, 02:37
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