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

Question

In a certain sequence, term  1   = 64, and for all n > 1, term  n   = (2^n)(term  n-1  ) 
What is the value of term  11  /term  8   ?

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

*kudos for all correct solutions

Hero8888 · Accepted Answer

A(n)=2^n*A(n-1) , hence

A(11)=2^{11} * A(10) ,
 A(10)=2^{10} * A(9) ,
 A(9)=2^9*A(8)

So  A(11)=2^{11}*2^{10}*2^9*A(8)=2^{30}*A(8) 
 A(11)/A(8)=2^{30}

Answer E.

Tuanguyen248 · Answer

I'm not sure but it seems like this

In a certain sequence,  term1 = 64 , and for all n > 1, term n  =  2^n   (n-1) ]

If so we got  \frac{term11}{term8}= \frac{2^{11}*2^{10}*2^9....2^2*64}{2^8*2^7*...2^2*64} = 2^{11}*2^{10}*2^9 = 2^{30}  => E

BrentGMATPrepNow · Answer

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

term 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

gmatophobia · Answer

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

Nikkz99 · Answer

Is solving within 3.5 min fine? Unable to solve within 2-2.5 min : (

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

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

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