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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. 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.

Are we assuming that, since j comes after i in the alphabet, that j is equivalent to i+1? If so, it is much clearer to write n*a_n=(n+1)*a_(n+1) because i and j could have any relationship

We are told that "\(i*a_i=j*a_j\) for any pair of positive integers \((i, j)\)", so it has nothing to do whether j comes after i and we are not assuming that j=i+1. _________________

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: \(2a_{100}=\frac{100}{99}a_{100}+\frac{100}{98}a_{100}\).

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.

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.

The new edition of question is clearer than the previous one.

I believe there is a typo in explanation of I. It should be Not True.

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.

Are we assuming that, since j comes after i in the alphabet, that j is equivalent to i+1? If so, it is much clearer to write n*a_n=(n+1)*a_(n+1) because i and j could have any relationship

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