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
OA E
Normally for this kind of the questions we need to find the pattern and solve it.
But I don't get any pattern after seeing the question.
Regards,
Uva.
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Identifying the patern
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theCEO
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Those answer choices look weird! Could you confirm that this question does not have any typo?Uva@90 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
OA E
Normally for this kind of the questions we need to find the pattern and solve it.
But I don't get any pattern after seeing the question.
Regards,
Uva.
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Hi Uva@90,
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
The answer choices appear to be missing the necessary notation, but it appears that 289 should probably be written as 2^89.
Final Answer: E
GMAT assassins aren't born, they're made,
Rich
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
The answer choices appear to be missing the necessary notation, but it appears that 289 should probably be written as 2^89.
Final Answer: E
GMAT assassins aren't born, they're made,
Rich
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GMATGuruNY
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As Rich has noted, the answer choices should read as I've posted them above.In the sequence 1, 2, 2, ..., An, An = A(n-1) "¢ A(n-2). The value of Aâ‚�₃ is how many times the value of Aâ‚�â‚�?
(A) 2
(B) 2³
(C) 2³²
(D) 2��
(E) 2��
Write it out and LOOK FOR A PATTERN.
Phrase the values in the sequence as the answer choices are phrased: in terms of POWERS OF 2.
A� = 1 = 2�.
A₂ = 2¹.
A₃ = A₂ * A� = 2¹ * 2� = 2¹.
A₄ = A₃ * A₂ = 2*2 = 2².
A₅ = A₄ * A₃ = 2² * 2 = 2³.
A₆ = A₅ * A₄ = 2³ * 2² = 2�.
A₇ = A₆ * A₅ = 2� * 2� = 2�.
Notice the pattern exhibited by the resulting exponents:
Each exponent is equal to the SUM OF THE TWO PRECEDING EXPONENTS.
To illustrate:
The exponent for A₆ (5) is equal to the sum of the exponents for A₅ and A₄ (3 and 2).
The exponent for A₇ (8) is equal to the sum of the exponents for A₆ and A₅ (5 and 3).
Thus, the sequence will proceed as follows:
Exponent for A₈ = sum of the exponents for A₇ and A₆ = 8+5 = 13.
Exponent for A₉ = sum of the exponents for A₈ and A₇ = 8+5 = 13 + 8 = 21.
Exponent for A�₀ = sum of the exponents for A₉ and A₈ = 21 + 13 = 34.
Exponent for A�� = sum of the exponents for A�₀ and A₉ = 34 + 21 = 55.
Exponent for A�₂ = sum of the exponents for A�� and A�₀ = 855+34 = 89.
Exponent for A�₃ = sum of the exponents for A�₂ and A�� = 89 + 55 = 144.
Thus:
A�₃/A��= 2¹��/2�� = 2��.
The correct answer is E.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
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