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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
_________________
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 
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
A. I only
B. II only
C. I and III only
D. II and III only
E. I, II, and III
I.
II.
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
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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 \(a_1=\text{positive integer}\). Thus, \(1*a_1=2*a_2=3*a_3=4*a_4=5*a_5=...=\text{positive integer}\).
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 \(a_1=\text{positive integer}=n*a_n\), then \(a_n=\frac{\text{positive integer}}{n}=\text{positive number}\). Therefore, this option is always true.
Answer: D
Hope it's clear.
_________________
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 \(a_1=\text{positive integer}\). Thus, \(1*a_1=2*a_2=3*a_3=4*a_4=5*a_5=...=\text{positive integer}\).
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 \(a_1=\text{positive integer}=n*a_n\), then \(a_n=\frac{\text{positive integer}}{n}=\text{positive number}\). Therefore, this option is always true.
Answer: D
Hope it's clear.
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LalaB
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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General Discussion
GMAT 1: 740 Q51 V38
Posts: 74
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
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
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Last visit: 17 Jul 2016
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Concentration: Finance
Schools: HEC '15 (A)
GMAT 1: 690 Q47 V38
Bunuel
Bunuel,
I didnt get this part. I seem to misunderstood the q.stem
could u please clarify it?
vinnik
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)
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Bunuel
could you please go thru this part one more time?
Cant get it
could you please go thru this part one more time?
Cant get it
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Galiya
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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Posts: 143
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?
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?
voodoochild
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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CleanSlate
Joined: 22 Oct 2012
Last visit: 13 Feb 2022
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GPA: 3.8
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Bunuel
can you please explain me option A. i am totally confused with it
FTG
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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Bunuel
Hi
I did not follow the move in bold. Can someone pls. explain a little more?
ronr34
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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pakasaip
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Bunuel
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
pakasaip
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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vinnik
Right..only looks complicated(that's how you should think first for some confidence
)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)
Posts: 4,849
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)
Devmitra Sen
GMAT SME
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)
Devmitra Sen
GMAT SME
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
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
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!
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!
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