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# In a certain sequence, term1 = 64, and for all

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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

Cheers,
Brent
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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
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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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