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# M17-25

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Intern
Joined: 07 Jul 2010
Posts: 6
Schools: AGSM '15
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Kudos [?]: 0 [0], given: 4

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11 Jun 2012, 23:51
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

Math Expert
Joined: 02 Sep 2009
Posts: 38958
Followers: 7748

Kudos [?]: 106416 [1] , given: 11626

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18 Sep 2012, 01:16
1
KUDOS
Expert's post
dandarth1 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

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.

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.
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 38958
Followers: 7748

Kudos [?]: 106416 [1] , given: 11626

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10 Feb 2013, 03:19
1
KUDOS
Expert's post
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

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.

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.
_________________
Intern
Status: ISB 14...:)
Joined: 26 May 2012
Posts: 30
Location: India
Concentration: Strategy
Schools: ISB '14 (A)
GMAT 1: 750 Q51 V39
GPA: 3.62
WE: Engineering (Energy and Utilities)
Followers: 1

Kudos [?]: 35 [0], given: 11

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12 Jun 2012, 01:33
i*A(i) = j*A(j)

Take i=1,
A(j) = A(1)/j
so,
A(2) = A(1)/2
A(3) = A(1)/3
.
.
.
The series is A(1), A(1)/2, A(1)/3...

I 2A(1)/100 = A(1)/99 + A(1)/98
2/100 = 1/99 + 1/98, this is definitely not true.

II IF A(1)=1, then all other numbers in the series will not be integers,
so II is possible.

III. Since A(1) is positive integer,
remaining numbers in the series will be positive only. This is always true.

So, II and III CAN be true.
Math Expert
Joined: 02 Sep 2009
Posts: 38958
Followers: 7748

Kudos [?]: 106416 [0], given: 11626

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12 Jun 2012, 05:27
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

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.

Hope it's clear.
_________________
Manager
Joined: 12 May 2012
Posts: 83
Location: India
Concentration: General Management, Operations
GMAT 1: 650 Q51 V25
GMAT 2: 730 Q50 V38
GPA: 4
WE: General Management (Transportation)
Followers: 2

Kudos [?]: 103 [0], given: 14

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12 Jun 2012, 10:12
Bunuel wrote:
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.

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.
Intern
Joined: 23 Aug 2012
Posts: 13
Followers: 0

Kudos [?]: 1 [0], given: 8

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17 Sep 2012, 21:14
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

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.

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
Manager
Status: K... M. G...
Joined: 22 Oct 2012
Posts: 51
GMAT Date: 08-27-2013
GPA: 3.8
Followers: 0

Kudos [?]: 9 [0], given: 118

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09 Feb 2013, 06:15
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

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.

Hope it's clear.

can you please explain me option A. i am totally confused with it
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Posts: 13
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07 Nov 2013, 21:41
Bunuel wrote:
COULD be true, not MUS be true).

'MUS' typo is even in the answer to this question (M17-25)
Re: M17-25   [#permalink] 07 Nov 2013, 21:41
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# M17-25

Moderator: Bunuel

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