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

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Senior Manager
Joined: 31 Oct 2011
Posts: 325
If a1, a2, a3, ..., an, ... is a sequence such that an = 2n [#permalink]

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28 Mar 2012, 01:57
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71% (02:01) correct 29% (01:21) wrong based on 228 sessions

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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
[Reveal] Spoiler: OA
Math Expert
Joined: 02 Sep 2009
Posts: 39607
Re: If a1, a2, a3, ..., an, ... is a sequence such that an = 2n [#permalink]

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28 Mar 2012, 02:12
1
KUDOS
Expert's post
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.

Hope it's clear.
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Posts: 262
Location: India
Concentration: Strategy, Operations
GMAT 1: 520 Q42 V19
GMAT 2: 540 Q44 V21
WE: Information Technology (Computer Software)
Re: If a1, a2, a3, ..., an, ... is a sequence such that an = 2n [#permalink]

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31 Mar 2012, 11:08
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.

Hope it's clear.

But
if i & j are index numbers and in sequence J>I
M i correct?
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Math Expert
Joined: 02 Sep 2009
Posts: 39607
Re: If a1, a2, a3, ..., an, ... is a sequence such that an = 2n [#permalink]

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31 Mar 2012, 11:38
GMATD11 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.

Hope it's clear.

But
if i & j are index numbers and in sequence J>I
M i correct?

Not sure I understood your question, but i>j because it's given that i^2 > j^2.
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Manager
Joined: 28 Jul 2011
Posts: 238
Re: If a1, a2, a3, ..., an, ... is a sequence such that an = 2n [#permalink]

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31 Mar 2012, 22:21
Vote for B

Given
So we have set of consicative number

& n>=1

{2,4,6,8,10.....}

is ai>aj

(A) i + j = even
o + o = e
e + e = e

so,
if (i>j) then ai>aj
if(i<j) then ai<aj
if (i=j) then aai=aj

data not suffficient

(B)

i^2 > j^2

we know for sure that i > j as n>=1 - i & j cannot be -ve

data sufficient
Current Student
Joined: 06 Sep 2013
Posts: 1998
Concentration: Finance
Re: If a1, a2, a3, ..., an, ... is a sequence such that an = 2n [#permalink]

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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.

Hope it's clear.

Can't index numbers be decimals ever?
Math Expert
Joined: 02 Sep 2009
Posts: 39607
Re: If a1, a2, a3, ..., an, ... is a sequence such that an = 2n [#permalink]

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06 Oct 2013, 11:14
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.

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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Joined: 09 Sep 2013
Posts: 15926
Re: If a1, a2, a3, ..., an, ... is a sequence such that an = 2n [#permalink]

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29 Jan 2016, 06:35
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Joined: 04 Jun 2016
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GMAT 1: 750 Q49 V43
Re: If a1, a2, a3, ..., an, ... is a sequence such that an = 2n [#permalink]

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16 Jul 2016, 01:15
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

(1) i is odd and j is even.
Not Sufficient. We don't know whether the odd or the even is bigger in magnitude

(2) i^2 > j^2
Sufficient :- Since numbers are non negative it means there is no surprises of mistakenly squaring a smaller negative.
A bigger squared value means a bigger base value
so i>j

Sufficient

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Re: If a1, a2, a3, ..., an, ... is a sequence such that an = 2n   [#permalink] 16 Jul 2016, 01:15
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