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Re: In a certain sequence, term1 = 64, and for all [#permalink]
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GMATPrepNow 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


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

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Re: In a certain sequence, term1 = 64, and for all [#permalink]
When we write few terms in the series, we will see that the numbers in the power forms an arithmetic progression. Using that concept we can find the power for \(term_{11}\) and \(term_{8}\).

Once we obtain the power we can subtract the same to obtain the net power.

The working is shown in the attached image.

Option E
Attachments

Screenshot 2022-09-16 125234.png
Screenshot 2022-09-16 125234.png [ 105.6 KiB | Viewed 1579 times ]

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Re: In a certain sequence, term1 = 64, and for all [#permalink]
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Re: In a certain sequence, term1 = 64, and for all [#permalink]
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