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If the infinite sequence a1, a2, a3, ..., an, ..., each term

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If the infinite sequence a1, a2, a3, ..., an, ..., each term [#permalink]

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New post 16 May 2008, 03:50
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In the infinite sequence \(a_1\), \(a_2\), \(a_3\),...., \(a_n\), each term after the first is equal to twice the previous term. If \(a_5-a_2=12\), what is the value of \(a_1\)?

A. 4
B. 24/7
C. 2
D. 12/7
E. 6/7

[Reveal] Spoiler:
The sequence looks more like x,x^2,x^4,x^8,x^16

a5 = x^16
a2=x^2

x^16-x^2 = 12
x^2(x^8 -1 ) = 12

I'm lost here. Thanks


OPEN DISCUSSION OF THIS QUESTION IS HERE: if-the-infinite-sequence-a1-a2-a3-an-each-term-134617.html
[Reveal] Spoiler: OA

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Last edited by Bunuel on 25 Mar 2014, 08:48, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.

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Re: Infinite sequence [#permalink]

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New post 16 May 2008, 05:46
Is it 6/7 ?
the seq wil be x , 2x, 4x, 8x...........not x^2,x^4 etc

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Re: Infinite sequence [#permalink]

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New post 16 May 2008, 05:51
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alimad wrote:
In the infinite sequence, a1,a2,a3,,,,an, each term after the first is equal to twice the previous term. If a5-a2 =12, what is the value of a1?


The sequence looks more like x,x^2,x^4,x^8,x^16

a5 = x^16
a2=x^2

x^16-x^2 = 12
x^2(x^8 -1 ) = 12

I'm lost here. Thanks


The sequence is
x, 2*x, 2*(2*x), 2*(2*(2*x)) .....
i.e.

nth term = 2^(n-1)x

a5=2^4 * x
a2=2^2*x

=> a5-a2 = (16-4)x= 12x
thus, 12x=12
and the first term x=1

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Re: Infinite sequence [#permalink]

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New post 16 May 2008, 07:17
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alimad wrote:
In the infinite sequence, a1,a2,a3,,,,an, each term after the first is equal to twice the previous term. If a5-a2 =12, what is the value of a1?


The sequence looks more like x,x^2,x^4,x^8,x^16

a5 = x^16
a2=x^2

x^16-x^2 = 12
x^2(x^8 -1 ) = 12

I'm lost here. Thanks


x. 2x. 4x. 8x. 16x. 16x-2x=12 14x=12. x=6/7

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Re: Infinite sequence [#permalink]

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New post 16 May 2008, 07:18
anirudhoswal wrote:
alimad wrote:
In the infinite sequence, a1,a2,a3,,,,an, each term after the first is equal to twice the previous term. If a5-a2 =12, what is the value of a1?


The sequence looks more like x,x^2,x^4,x^8,x^16

a5 = x^16
a2=x^2

x^16-x^2 = 12
x^2(x^8 -1 ) = 12

I'm lost here. Thanks


The sequence is
x, 2*x, 2*(2*x), 2*(2*(2*x)) .....
i.e.

nth term = 2^(n-1)x

a5=2^4 * x
a2=2^2*x

=> a5-a2 = (16-4)x= 12x
thus, 12x=12
and the first term x=1


This cannot be correct.

Just try it. 1, 2, 4, 8, 16. 16-2 dsnt = 12.

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Re: Infinite sequence [#permalink]

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New post 16 May 2008, 19:00
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The OA is 6/7
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Re: In the infinite sequence, a1,a2,a3,,,,an, each term after [#permalink]

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New post 25 Mar 2014, 07:49
a5=2^4*x
a2=2^1*x

So a5-a2=16x-12x=14x
14x=12 => x=6/7

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Re: If the infinite sequence a1, a2, a3, ..., an, ..., each term [#permalink]

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New post 25 Mar 2014, 08:46
In the infinite sequence \(a_1\), \(a_2\), \(a_3\),...., \(a_n\), each term after the first is equal to twice the previous term. If \(a_5-a_2=12\), what is the value of \(a_1\)?

A. 4
B. 24/7
C. 2
D. 12/7
E. 6/7

The formula for calculating \(n_{th}\) term would be \(a_n=2^{n-1}*a_1\) . So:
\(a_5=2^4*a_1\);
\(a_2=2*a_1\);

Given: \(a_5-a_2=2^4*a_1-2*a_1=12\) --> \(2^4*a_1-2*a_1=12\) --> \(a_1=\frac{12}{14}=\frac{6}{7}\).

Answer: E.

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-the-infinite-sequence-a1-a2-a3-an-each-term-134617.html
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Re: If the infinite sequence a1, a2, a3, ..., an, ..., each term [#permalink]

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Re: If the infinite sequence a1, a2, a3, ..., an, ..., each term [#permalink]

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New post 30 Aug 2017, 11:27
Process elimination could work as well.
Answer 6/7

6/7, 12/7, 24/7, 48/7, 96/7

a5 - a2
96/7 - 12/7
12
Answer is E

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Re: If the infinite sequence a1, a2, a3, ..., an, ..., each term [#permalink]

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New post 02 Sep 2017, 07:08
alimad wrote:
In the infinite sequence \(a_1\), \(a_2\), \(a_3\),...., \(a_n\), each term after the first is equal to twice the previous term. If \(a_5-a_2=12\), what is the value of \(a_1\)?

A. 4
B. 24/7
C. 2
D. 12/7
E. 6/7


We can let a_1 = x, a_2 = 2x, a_3 = 4x, a_4 = 8x and a_5 = 16x. Thus:

16x - 2x = 12

14x = 12

x = 12/14 = 6/7

Answer: E
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Re: If the infinite sequence a1, a2, a3, ..., an, ..., each term   [#permalink] 02 Sep 2017, 07:08
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