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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

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.

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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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 \(a_1=positive \ integer\)).

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

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17 Nov 2015, 20:57

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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.

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.
_________________

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..

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