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In a certain sequence , term[color=red]1[/color] = 64

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In a certain sequence , term[color=red]1[/color] = 64 [#permalink] New post 13 Jun 2018, 06:30
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In a certain sequence, term1 = 64, and for all n > 1, termn = (2^n)(termn-1)
What is the value of term11/term8 ?

A) 2^3
B) 2^6
C) 2^9
D) 2^27
E) 2^30
[Reveal] Spoiler: OA

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Re: In a certain sequence , term[color=red]1[/color] = 64 [#permalink] New post 13 Jun 2018, 08:56
The answer is E?
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Re: In a certain sequence , term[color=red]1[/color] = 64 [#permalink] New post 13 Jun 2018, 09:21
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sam_ridhi wrote:
The answer is E?


Sorry, I didn't add the correct answer.
Yes, the answer is E.

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Brent
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Re: In a certain sequence , term[color=red]1[/color] = 64 [#permalink] New post 13 Jun 2018, 09:53
how's it E?
please explain.
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Re: In a certain sequence , term[color=red]1[/color] = 64 [#permalink] New post 13 Jun 2018, 09:55
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Sona4292 wrote:
how's it E?
please explain.


I'll post a solution in 2 days.
In the meantime, see you do with the question.

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Brent
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Re: In a certain sequence , term[color=red]1[/color] = 64 [#permalink] New post 14 Jun 2018, 00:15
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The first term of the sequence is given as 64 = 2^6
From the formula,
2nd term = \(2^2 * 2^6\)
3rd term = \(2^3 * 2^2 * 2^6\)
4th term = \(2^4 * 2^3 * 2^2 * 2^6\)

There is a pattern with 2^6 constant for every term and the power of 2 is raised consecutively from \(2^2 to 2^n\)

Hence, \(2^1^1 = 2^6 * 2^2 * 2^3 ...........*2^1^1\)
similarly \(2^8 = 2^6 * 2^2 * 2^3.........*2^8\)

when finding the ratio everything cancels leaving only \(2^9 * 2^10 * 2^1^1 = 2^3^0\)
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This is my response to the question and may be incorrect. Feel free to rectify any mistakes

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Re: In a certain sequence , term[color=red]1[/color] = 64 [#permalink] New post 15 Jun 2018, 08:27
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GreenlightTestPrep wrote:
In a certain sequence, term1 = 64, and for all n > 1, termn = (2^n)(termn-1)
What is the value of term11/term8 ?

A) 2^3
B) 2^6
C) 2^9
D) 2^27
E) 2^30


*kudos for all correct solutions[/quote]

Let's list a few terms and look for a pattern

term1 = 64 = 2^8
term2 = (2^8)(2^2)
term3 = (2^8)(2^2)(2^3)
term4 = (2^8)(2^2)(2^3)(2^4)
.
.
.
term8 = (2^8)(2^2)(2^3)(2^4)(2^5)(2^6)(2^7)(2^8)
.
.
.
term11 = (2^8)(2^2)(2^3)(2^4)(2^5)(2^6)(2^7)(2^8)(2^9)(2^10)(2^11)

So, term11/term8 = (2^8)(2^2)(2^3)(2^4)(2^5)(2^6)(2^7)(2^8)(2^9)(2^10)(2^11)/(2^8)(2^2)(2^3)(2^4)(2^5)(2^6)(2^7)(2^8)
= (2^9)(2^10)(2^11)
= 2^30

Answer: E

Cheers,
Brent
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Re: In a certain sequence , term[color=red]1[/color] = 64   [#permalink] 15 Jun 2018, 08:27
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In a certain sequence , term[color=red]1[/color] = 64

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