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Re: The sequence a1, a2, a3, ..., an, ... is such that i*a_i=j*aj for any [#permalink]
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i*A(i)=j*A(j) for any pair of positive integers means i*A(i) is constant for any i
So i * A(i) = C (C is a constant)
A(i)= C/i for any positive integers
A(1)=C is an integer
A(100)= C/100
A(98)= C/98
A(99)=C/99
I. 2 * A(100)= 2C/100= C/50
A(99)=C/99, A(98)=C/98
Because C is an positive integer so C can not be Zero. I is impossible because C/50 cannot equal C(1/99+1/98)
II. if C=1, A(n)=C/n so A(1) is the only integer
III. C is a positive integer so A(i)= C/i can not be negative
D is my answer.
By the way, I need 3 more kudos. So if I'm right, Kudo me please
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Re: The sequence a1, a2, a3, ..., an, ... is such that i*a_i=j*aj for any [#permalink]
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: The sequence a1, a2, a3, ..., an, ... is such that i*a_i=j*aj for any [#permalink]
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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: The sequence a1, a2, a3, ..., an, ... is such that i*a_i=j*aj for any [#permalink]
Bunuel
could you please go thru this part one more time?
Cant get it
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Re: The sequence a1, a2, a3, ..., an, ... is such that i*a_i=j*aj for any [#permalink]
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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: The sequence a1, a2, a3, ..., an, ... is such that i*a_i=j*aj for any [#permalink]
Bunuel,
I agree with (i) and (iii). However, I am not sure about (ii).

Why did you substitute a1 =1 ? If A(1) is the only integer => n=1; But how do we know that a1 = 1? a1 could be anything....a1=2 also holds good because there is only one number. Correct? Essentially, there is no A(2), A(3) etc.

Thoughts?
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Re: The sequence a1, a2, a3, ..., an, ... is such that i*a_i=j*aj for any [#permalink]
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voodoochild wrote:
Bunuel,
I agree with (i) and (iii). However, I am not sure about (ii).

Why did you substitute a1 =1 ? If A(1) is the only integer => n=1; But how do we know that a1 = 1? a1 could be anything....a1=2 also holds good because there is only one number. Correct? Essentially, there is no A(2), A(3) etc.

Thoughts?


The question asks "which of the following is possible" or which of the following COULD be true. So, we don't know that \(a_1=1\), but \(a_1\) COULD be 1 and in this case it would be the only integer in the sequence. So, II is certainly POSSIBLE.

Hope it's clear.
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Re: The sequence a1, a2, a3, ..., an, ... is such that i*a_i=j*aj for any [#permalink]
Bunuel wrote:
RuslanMRF 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. 2A(100)=A(99)+A(98)
II. A(1) is the only integer in the series
III. The series does not contain negative numbers

I only
II only
I and III only
II and III only
I, II, and III

Please, explain the the solution.


New edition of this question with a solution:

The sequence \(a_1\), \(a_2\), \(a_3\), ..., \(a_n\), ... 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 could be true?

I. \(2*a_{100}=a_{99}+a_{98}\)
II. \(a_1\) is the only integer in the sequence
III. The sequence does not contain negative numbers

A. I only
B. II only
C. I and III only
D. II and III only
E. I, II and III

Given that the 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. \(2a_{100}=a_{99}+a_{98}\) --> since \(100a_{100}=99a_{99}=98a_{98}\), 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 could be true.

II. \(a_1\) is the only integer in the sequence. 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 could be true.

III. The sequence does not contain negative numbers --> since 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.

Hope it's clear.



can you please explain me option A. i am totally confused with it
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Re: The sequence a1, a2, a3, ..., an, ... is such that i*a_i=j*aj for any [#permalink]
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FTG wrote:
Bunuel wrote:
RuslanMRF 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. 2A(100)=A(99)+A(98)
II. A(1) is the only integer in the series
III. The series does not contain negative numbers

I only
II only
I and III only
II and III only
I, II, and III

Please, explain the the solution.


New edition of this question with a solution:

The sequence \(a_1\), \(a_2\), \(a_3\), ..., \(a_n\), ... 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 could be true?

I. \(2*a_{100}=a_{99}+a_{98}\)
II. \(a_1\) is the only integer in the sequence
III. The sequence does not contain negative numbers

A. I only
B. II only
C. I and III only
D. II and III only
E. I, II and III

Given that the 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. \(2a_{100}=a_{99}+a_{98}\) --> since \(100a_{100}=99a_{99}=98a_{98}\), 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 could be true.

II. \(a_1\) is the only integer in the sequence. 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 could be true.

III. The sequence does not contain negative numbers --> since 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.

Hope it's clear.



can you please explain me option A. i am totally confused with 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: The sequence a1, a2, a3, ..., an, ... is such that i*a_i=j*aj for any [#permalink]
Bunuel wrote:
ksharma12 wrote:
Series is such that for any pair of positive integers . If is a positive integer, which of the following is possible?

I.

II. is the only integer in the series

III. The series does not contain negative numbers


I only
II only
I and III only
II and III only
I, II, and III

I have no idea whats going on here? Detailed explanation is appreciated.


A set 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 (note that the question is which can be true, not must 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.


Hi
I did not follow the move in bold. Can someone pls. explain a little more?
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Re: The sequence a1, a2, a3, ..., an, ... is such that i*a_i=j*aj for any [#permalink]
Expert Reply
ronr34 wrote:
Bunuel wrote:
ksharma12 wrote:
Series is such that for any pair of positive integers . If is a positive integer, which of the following is possible?

I.

II. is the only integer in the series

III. The series does not contain negative numbers


I only
II only
I and III only
II and III only
I, II, and III

I have no idea whats going on here? Detailed explanation is appreciated.


A set 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 (note that the question is which can be true, not must 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.


Hi
I did not follow the move in bold. Can someone pls. explain a little more?


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: The sequence a1, a2, a3, ..., an, ... is such that i*a_i=j*aj for any [#permalink]
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: The sequence a1, a2, a3, ..., an, ... is such that i*a_i=j*aj for any [#permalink]
Expert Reply
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: The sequence a1, a2, a3, ..., an, ... is such that i*a_i=j*aj for any [#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


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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Re: The sequence a1, a2, a3, ..., an, ... is such that i*a_i=j*aj for any [#permalink]
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Solution:

Lets first understand i x Ai = j x Aj

Ai is a positive integer.

Notice that if i = 1 and j = 2, A(1) = 2*A(2)

=>A(2) = A(1)/2

Again if, i = 1 and j = 3, A(1) = 3*A(3) [ i is not necessarily equal to j]

So A(3) = A(1)/3

=> A(1), A(1)/2, A(1)/3, A(1)/4 can represent the series

Clearly A(1) is the ONLY integer and II is true (Eliminate A and C)

Also, since A(1) is a positive integer, no other numbers following it can be negative. So III is true too.
(Eliminate B)

Now, 2*A(100) = A(99) + A(98)

=>2*A(1)/100 = A(1)/99 + A(1)/98 (cancel A(1) here)

=>2/100 = 1/99 + 1/98

Not true (Incorrect conceptually)
So eliminate E.

(option D)

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Re: The sequence a1, a2, a3, ..., an, ... is such that i*a_i=j*aj for any [#permalink]
100a100 = 99a99 = 98a98

100a100/50 = 2a100

99a99/100 + 98a98/100 = 2a100

This is not equal to

2a100 = a99 + a98

I could not be true.

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Re: The sequence a1, a2, a3, ..., an, ... is such that i*a_i=j*aj for any [#permalink]
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