In the sequence 1, 2, 2, …, an, …, an = an-1 • an-2.

Question

In the sequence 1, 2, 2, …,  a_n , …,  a_n = a_{n-1}* a_{n-2} . The value of  a_{13}  is how many times the value of  a_{11} ?

(A) 2
(B) 2^3
(C) 2^32
(D) 2^64
(E) 2^89

Disclaimer: I have used the Search Box Before Posting. I used the first sentence of the question or a string of words exactly as they show up in the question below for my search. I did not receive an exact match for my question. 
Source: Veritas Prep; Book 04
Chapter: Homework
Topic: Algebra
Question: 93
Question: Page 226
Solution: PDF Page 17 of 18
Edition: Third

My Question: Please provide an explanation on how to arrive at the answer.

Bunuel · Accepted Answer

The question should read:
 In the sequence 1, 2, 2, …,  a_n , …,  a_n = a_{n-1}* a_{n-2} . The value of  a_{13}  is how many times the value of  a_{11} ?  
(A) 2
(B) 2^3
(C) 2^32
(D) 2^64
(E) 2^89

For such kind of questions it's almost always a good idea to write down first terms:
 a_1=1=2^0 
 a_2=2=2^1 
 a_3=a_2*a_1=1*2=2^1 
 a_4=a_3*a_2=2*2=2^2 
 a_5=a_4*a_3=4*2=2^3 
 a_6=a_5*a_4=8*4=2^5 
 a_7=a_6*a_5=32*8=2^8

If you notice exponents form Fibonacci sequence: {0, 1, 1, 2, 3, 5, 8, ...} (Fibonacci sequence is a sequence where each subsequent number is the sum of the previous two)

So, it will continue as follows: {0, 1, 1, 2, 3, 5, 8, 5+8= 13 , 8+13= 21 , 13+21= 34 , 21+34= 55 , 34+55= 89 , 55+89= 144 , ...}

From above we have that  a_{11}=2^{55}  and  a_{13}=2^{144} .

\frac{a_{13}}{a_{11}}=\frac{2^{144}}{2^{55}}=2^{89}

Answer: E.

baitthegmat · Answer

Damn It! The original post of question drove me crazy for good 30 min. I tried both ways, just trying to figure out the pattern and calculate for a12, and solving for every value in the sequence until a13. Still could not get even close to the answer choices. I almost always just keep the page scrolled enough so that I can read only the question, and not the answers or any explanation by mistake.

It was a blunder in this time. After spending all that time, I figured out there were typos, and I calculated correct answer in well within 1.5 min in my first go. Thanks Bunuel! You just saved me. Phewwwww..

It was a blunder in this time. After spending all that time, I figured out there were typos, and I calculated correct answer in well within 1.5 min in my first go. Thanks Bunuel! You just saved me. Phewwwww..

EMPOWERgmatRichC · Answer

Hi All,

The pattern in this question is rarer than the ones that you'll likely see on sequence questions on the Official GMAT - the pattern is based on "2 raised to a power"...

Since the first two terms are 1 and 2, and we're told to MULTIPLY the prior 2 terms in the sequence to get the next term in the sequence, the next few terms are...

3rd term = 2 = 2^1 
4th term = 4 = 2^2 
5th term = 8 = 2^3 
6th term = 32 = 2^5 
7th term = 256 = 2^8

From here, the pattern can redefined as "add up the EXPONENTS of the prior 2 terms"; in this way, you can map out the remaining terms in the sequence much faster...

8th term = 2^13 
9th term = 2^21 
10th term = 2^34 
11th term = 2^55 
12th term = 2^89 
13th term = 2^144

We're essentially asked for the value of (13th term)/(11th term)....

(2^144)/(2^55) = 2^89

Final Answer:  E

GMAT assassins aren't born, they're made, 
Rich

legendinthewomb · Answer

There isn't a typo in the question.

Kinshook · Answer

Asked: In the sequence 1, 2, 2, …,  a_n , …,  a_n = a_{n-1}* a_{n-2} . The value of  a_{13}  is how many times the value of  a_{11} ?

a_n = a_{n-1}* a_{n-2} 
Sequence =  {1,2,2,4,2^3,2^5,2^8,2^13,2^21,2^34,2^55,2^89,2^144...}

a_{13} = 2^{144} 
 a_{11} = 2^{55} 
 a_{13} = a^{11} * 2^{144-55} 
 a_{13} = a^{11} * 2^{89}

IMO E

bumpbot · Answer

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In the sequence 1, 2, 2, …, an, …, an = an-1 • an-2.

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