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If a1, a2, a3, ..., an, ... is a sequence such that an = 2n

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eybrj2
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If a1, a2, a3, ..., an, ... is a sequence such that an = 2n [#permalink] New post   28 Mar 2012, 01:57
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If \(a_1\), \(a_2\), \(a_3\), ..., \(a_n\), ... is a sequence such that \(a_n = 2n\) for all \(n \geq 1\), is \(a_i\) greater than \(a_j\)?

(1) i is odd and j is even.

(2) \(i^2 > j^2\)


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Bunuel
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Re: If a1, a2, a3, ..., an, ... is a sequence such that an = 2n [#permalink] New post   28 Mar 2012, 02:12
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eybrj2 wrote:
If a1, a2, a3, ..., an, ... is a sequence such that an = 2n for all n>= 1, is ai greater than aj?

(1) i is add and j is even.

(2) i^2 > j^2


Since given that \(a_n = 2n\), for all \(n\geq{1}\) then:
\(a_1=2*1=2\);
\(a_2=2*2=4\);
\(a_3=2*3=6\);
\(a_4=2*4=8\);
...

Basically we have a sequence of positive even numbers. Question asks whether \(a_i>a_j\)? So, it basically asks whether \(i>j\)?

(1) i is add and j is even. Not sufficient.

(2) i^2 > j^2 --> since \(i\) and \(j\) are both positive integers (they represent index numbers) then \(i>j\). Sufficient.

Answer: B.

Hope it's clear.
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Re: If a1, a2, a3, ..., an, ... is a sequence such that an = 2n [#permalink] New post   06 Oct 2013, 11:04
Bunuel wrote:
eybrj2 wrote:
If a1, a2, a3, ..., an, ... is a sequence such that an = 2n for all n>= 1, is ai greater than aj?

(1) i is add and j is even.

(2) i^2 > j^2


Since given that \(a_n = 2n\), for all \(n\geq{1}\) then:
\(a_1=2*1=2\);
\(a_2=2*2=4\);
\(a_3=2*3=6\);
\(a_4=2*4=8\);
...

Basically we have a sequence of positive even numbers. Question asks whether \(a_i>a_j\)? So, it basically asks whether \(i>j\)?

(1) i is add and j is even. Not sufficient.

(2) i^2 > j^2 --> since \(i\) and \(j\) are both positive integers (they represent index numbers) then \(i>j\). Sufficient.

Answer: B.

Hope it's clear.


Can't index numbers be decimals ever?
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Bunuel
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Re: If a1, a2, a3, ..., an, ... is a sequence such that an = 2n [#permalink] New post   06 Oct 2013, 11:14
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jlgdr wrote:
Bunuel wrote:
eybrj2 wrote:
If a1, a2, a3, ..., an, ... is a sequence such that an = 2n for all n>= 1, is ai greater than aj?

(1) i is add and j is even.

(2) i^2 > j^2


Since given that \(a_n = 2n\), for all \(n\geq{1}\) then:
\(a_1=2*1=2\);
\(a_2=2*2=4\);
\(a_3=2*3=6\);
\(a_4=2*4=8\);
...

Basically we have a sequence of positive even numbers. Question asks whether \(a_i>a_j\)? So, it basically asks whether \(i>j\)?

(1) i is add and j is even. Not sufficient.

(2) i^2 > j^2 --> since \(i\) and \(j\) are both positive integers (they represent index numbers) then \(i>j\). Sufficient.

Answer: B.

Hope it's clear.


Can't index numbers be decimals ever?


n in \(a_n\) shows which term is \(a_n\) in sequence so it cannot be a decimal.
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beattheheat
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Re: If a1, a2, a3, ..., an, ... is a sequence such that an = 2n [#permalink] New post   19 Dec 2014, 10:37
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If a1, a2, a3, . . . , an, . . . is a sequence such that a_n=2n for all n ≥ 1, is a_i greater than a_j ?

(1) i is odd and j is even
(2) i^2 > j^2

Solution:
Here each term is twice of previous term. Need to identify whether i > j?

Statement 1: i is odd and j is even
Doesn't provide any relation between i and j - Insufficient

Statement 2: i^2 > j^2
Since a_i and a_j are terms of sequence, i and j must be positive integers.
Thus i must be greater than j and hence Statement 2 alone is Sufficient.

Answer B
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chisichei
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Re: If a1, a2, a3, ..., an, ... is a sequence such that an = 2n [#permalink] New post   18 Mar 2018, 02:06
Hi Bunuel please the question does not tell you if i or j is ≥ 1 and that is why i chose E.
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Re: If a1, a2, a3, ..., an, ... is a sequence such that an = 2n [#permalink] New post   18 Mar 2018, 05:51
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chisichei wrote:
Hi Bunuel please the question does not tell you if i or j is ≥ 1 and that is why i chose E.


i and j are index numbers indicating which position a number has in the sequence. The sequence starts with a1, so both i and j must be more than or equal to 1.
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FlyingCycle
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Re: If a1, a2, a3, ..., an, ... is a sequence such that an = 2n [#permalink] New post   19 Jun 2021, 08:07
Bunuel

I did this way:

i^2 - j^2 > 0

(i+j)(i-j)>0

so either both are negative or both are positive.

since all numbers are more than 1 (positive), therefore, i & j value must be positive. Which means i + j is positive and then i-j value must be positive too, so i > j

can I solve this way?
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Re: If a1, a2, a3, ..., an, ... is a sequence such that an = 2n [#permalink] New post   05 Jan 2024, 13:41
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Re: If a1, a2, a3, ..., an, ... is a sequence such that an = 2n [#permalink]
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