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In a certain sequence , term[color=red]1[/color] = 64 [#permalink]
13 Jun 2018, 06:30
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In a certain sequence, term 1 = 64, and for all n > 1, term n = (2^n)(term n1) What is the value of term 11/term 8 ? A) 2^3 B) 2^6 C) 2^9 D) 2^27 E) 2^30
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Re: In a certain sequence , term[color=red]1[/color] = 64 [#permalink]
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]
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sam_ridhi wrote: The answer is E? Sorry, I didn't add the correct answer. Yes, the answer is E. Cheers, Brent
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Re: In a certain sequence , term[color=red]1[/color] = 64 [#permalink]
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]
13 Jun 2018, 09:55
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. Cheers, Brent
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Re: In a certain sequence , term[color=red]1[/color] = 64 [#permalink]
14 Jun 2018, 00:15
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The first term of the sequence is given as 64 = 2^6From 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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Re: In a certain sequence , term[color=red]1[/color] = 64 [#permalink]
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GreenlightTestPrep wrote: In a certain sequence, term1 = 64, and for all n > 1, termn = (2^n)(termn1) 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 patternterm 1 = 64 = 2^6 term 2 = (2^6)(2^2) term 3 = (2^6)(2^2)(2^3) term 4 = (2^6)(2^2)(2^3)(2^4) . . . term 8 = (2^6)(2^2)(2^3)(2^4)(2^5)(2^6)(2^7)(2^8). . . term 11 = (2^6)(2^2)(2^3)(2^4)(2^5)(2^6)(2^7)(2^8)(2^9)(2^10)(2^11) So, term11/term8 = (2^6)(2^2)(2^3)(2^4)(2^5)(2^6)(2^7)(2^8)(2^9)(2^10)(2^11) /(2^6)(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]
20 Nov 2018, 16:12
GreenlightTestPrep wrote: GreenlightTestPrep wrote: In a certain sequence, term1 = 64, and for all n > 1, termn = (2^n)(termn1) 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 Let's list a few terms and look for a patternterm 1 = 64 = 2^8 term 2 = (2^8)(2^2) term 3 = (2^8)(2^2)(2^3) term 4 = (2^8)(2^2)(2^3)(2^4) . . . term 8 = (2^8)(2^2)(2^3)(2^4)(2^5)(2^6)(2^7)(2^8). . . term 11 = (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[/quote] 64=2^8 or 2^6



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Re: In a certain sequence , term[color=red]1[/color] = 64 [#permalink]
20 Nov 2018, 16:21
Good catch! Rookie mistake on my part. 64 = 2^6 (not 2^8) That said, the two expressions with 2^8 (or 2^6) cancel each other out. So, fortunately, the solution still worked out Cheers and thanks for the heads up! I have edited my answer accordingly. Cheers, Brent
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Re: In a certain sequence , term[color=red]1[/color] = 64
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