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

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:

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

There isn't a typo in the question.

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

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