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

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In the sequence 1, 2, 2, …, an, …, an = an-1 • an-2.  [#permalink]

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New post Updated on: 26 Jul 2013, 12:30
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65% (03:01) correct 35% (02:42) wrong based on 358 sessions

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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.

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Originally posted by hb on 26 Jul 2013, 12:02.
Last edited by hb on 26 Jul 2013, 12:30, edited 2 times in total.
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Re: In the sequence 1, 2, 2, …, an, …, an = an-1 • an-2.  [#permalink]

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New post 26 Jul 2013, 12:26
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hb wrote:
In the sequence 1, 2, 2, …, an, …, an = an-1 • an-2. The value of a13 is how many times the value of a11?

(A) 2
(B) 23
(C) 232
(D) 264
(E) 289


Comments: The n, n-1, n-2, 13, 11 mentioned in the question are subscripts above. I could not figure out how to show them as subscript while writing the formula here.

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.


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.
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Re: In the sequence 1, 2, 2, …, an, …, an = an-1 • an-2.  [#permalink]

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New post 25 Sep 2013, 11:16
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..
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Re: In the sequence 1, 2, 2, …, an, …, an = an-1 • an-2.  [#permalink]

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New post 20 Feb 2018, 16:20
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

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Re: In the sequence 1, 2, 2, …, an, …, an = an-1 • an-2.   [#permalink] 20 Feb 2018, 16:20
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