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# m17 q25

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Intern
Joined: 10 Jun 2010
Posts: 8

Kudos [?]: 3 [0], given: 4

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06 Jul 2010, 07:30
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

a. I only
b. II only
c. I and III only
d. II and III only
e. I, II, and III

[Reveal] Spoiler:
II is possible. If $$A(1)=1$$ then $$A(k) = \frac{1}{k}$$ . There are no more integers in the sequence except $$A(1)$$ .
III is possible as well. Same example applies.
I is not possible. Because $$i*A(i) = (i + 1)A(i + 1)$$$$A(i + 1) = \frac{i}{i + 1} A(i)$$ which is less then $$A(i)$$. This means that this sequence is a decreasing sequence in which every subsequent element is smaller than its predecessor. Thus, $$A(100) + A(100) \lt A(99) + A(98)$$
The correct answer is D.

Can someone please explain this?
I have no clue what data is provided and what answer is expected in this question

Kudos [?]: 3 [0], given: 4

Math Expert
Joined: 02 Sep 2009
Posts: 42559

Kudos [?]: 135309 [2], given: 12686

Re: m17 q25 [#permalink]

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08 Jul 2010, 18:01
2
KUDOS
Expert's post
yeahwill 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

a. I only
b. II only
c. I and III only
d. II and III only
e. I, II, and III

[Reveal] Spoiler:
II is possible. If $$A(1)=1$$ then $$A(k) = \frac{1}{k}$$ . There are no more integers in the sequence except $$A(1)$$ .
III is possible as well. Same example applies.
I is not possible. Because $$i*A(i) = (i + 1)A(i + 1)$$$$A(i + 1) = \frac{i}{i + 1} A(i)$$ which is less then $$A(i)$$. This means that this sequence is a decreasing sequence in which every subsequent element is smaller than its predecessor. Thus, $$A(100) + A(100) \lt A(99) + A(98)$$
The correct answer is D.

Can someone please explain this?
I have no clue what data is provided and what answer is expected in this question

This question was posted in PS forum. Below is my post from there:

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.

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.
_________________

Kudos [?]: 135309 [2], given: 12686

Manager
Joined: 16 Feb 2011
Posts: 195

Kudos [?]: 246 [0], given: 78

Schools: ABCD
Re: m17 q25 [#permalink]

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13 Jul 2012, 12:22
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?

Kudos [?]: 246 [0], given: 78

Math Expert
Joined: 02 Sep 2009
Posts: 42559

Kudos [?]: 135309 [0], given: 12686

Re: m17 q25 [#permalink]

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13 Jul 2012, 12:28
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.
_________________

Kudos [?]: 135309 [0], given: 12686

Re: m17 q25   [#permalink] 13 Jul 2012, 12:28
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# m17 q25

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