The sequence a1, a2, a3, ..., an, ... is such that i*a_i=j*aj for any

Question

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

M17-25

Series  https://gmatclub.com/cgi-bin/mimetex.cgi?A%28n%29  is such that  https://gmatclub.com/cgi-bin/mimetex.cgi?i%2AA%28i%29%20%3D%20j%2AA%28j%29  for any pair of positive integers  https://gmatclub.com/cgi-bin/mimetex.cgi?%28i%2C%20j%29  . If  https://gmatclub.com/cgi-bin/mimetex.cgi?A%281%29  is a positive integer, which of the following is possible?

I.  https://gmatclub.com/cgi-bin/mimetex.cgi?2A%28100%29%20%3D%20A%2899%29%20%2B%20A%2898%29

II.  https://gmatclub.com/cgi-bin/mimetex.cgi?A%281%29  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

Bunuel · Accepted Answer

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 , ... has the following properties:  i*a_i=j*a_j  and  a_1=	ext{positive integer} . Thus,  1*a_1=2*a_2=3*a_3=4*a_4=5*a_5=...=	ext{positive integer} .
 
We need to determine which of the given options can occur (note that the question asks which of the following  COULD be true, not MUST 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} . Divide by  a_{100} :  2=\frac{100}{99}+\frac{100}{98} , which is not true. Hence this option  could NOT be true .
 
II.  a_1  is the only integer in the sequence. If  a_1=1 , then all other terms will be non-integers, because in this case, we would have  a_1=1=2a_2=3a_3=... , which leads to  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. Given that  a_1=	ext{positive integer}=n*a_n , then  a_n=\frac{	ext{positive integer}}{n}=	ext{positive number} . Therefore, this option is  always true .

Answer: D

Hope it's clear.

Bunuel · Answer

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.

DavidArchuleta · Answer

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

LalaB · Answer

Bunuel,

I didnt get this part. I seem to misunderstood the q.stem

could u please clarify it?

KarishmaB · Answer

First thing I want to understand is this relation: i*A(i) = j*A(j) for any pair of positive integers. I will take examples to understand it.

When i = 1 and j = 2, A(1) = 2*A(2)
So A(2) = A(1)/2

When i = 1 and j = 3, A(1) = 3*A(3)
So A(3) = A(1)/3

I see it now. The series is: A(1), A(1)/2, A(1)/3, A(1)/4 and so on...

II and III are easily possible. We can see that without any calculations.

II. A(1) is the only integer in the series 
If A(1) = 1, then series becomes 1, 1/2, 1/3, 1/4 ... all fractions except A(1)

III. The series does not contain negative numbers 
Again, same series as above applies. In fact, since A(1) is a positive integer, this must be true.

I. 2*A(100) = A(99) + A(98) 
2*A(1)/100 = A(1)/99 + A(1)/98 (cancel A(1) from both sides)
2/100 = 1/99 + 1/98
Not true hence this is not possible

Answer (D)

galiya · Answer

Bunuel
could you please go thru this part one more time?
Cant get it

Bunuel · Answer

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.

voodoochild · Answer

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?

Bunuel · Answer

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.

CleanSlate · Answer

can you please explain me option A. i am totally confused with it

Bunuel · Answer

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.

ronr34 · Answer

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

Bunuel · Answer

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.

pakasaip · Answer

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

Bunuel · Answer

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.

ShashankDave · Answer

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)

CrackverbalGMAT · Answer

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)

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)

Devmitra Sen
GMAT SME

Bambi2021 · Answer

100a100 = 99a99 = 98a98

100a100/50 = 2a100

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

This is not equal to

2a100 = a99 + a98

I could not be true.

Posted from my mobile device

Lewy · Answer

3I have appreciated the invaluable insights gained in the various approaches used to tackle this problem. As such, I also want to give my feedback for anyone coming across this and feeling stuck.
Here we go!

a1, a2, a3 ..... an are the terms of the sequence and the relationship between any 2 such terms is such that i∗ai=j∗aj. What this means is that a term multiplied by its index is equal to another term multiplied by its index. That's all we should care about for now.
We can in essence, compare a1 to any term in the sequence and express such term in terms of a1:
 Given, i∗ai = j∗aj
 aj = (i * ai)/j, and since i =1
 aj = a1/j
 I. 2∗a100=a99+a98, aj = a100, a99, a98 substituting with a1/j becomes, 2*a1/100 = a1/99+a1/98.
 If we let a1 = 1 you can see that the expression can't hold to be true.
 For II. a1 is the only integer in the sequence
 As shown prior, aj = a1/j , taking the first 3 indexes i.e 1, 2, 3 for j and a1 = 1, we can see that aj = 1/1, 1/2, 1/3 and as a result this statement is true
 For III. The sequence does not contain negative numbers
 Since a1 has been designated a +ve integer, all subsequent terms in the sequence will always have +ve values as shown in II above.

I hope this has helped clarify any misunderstood logic in the quiz. Thank you!

The sequence a1, a2, a3, ..., an, ... is such that ia_i=jaj for any

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The sequence a1, a2, a3, ..., an, ... is such that i*a_i=j*aj for any

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The sequence a1, a2, a3, ..., an, ... is such that ia_i=jaj for any