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In the finite sequence of positive integers A1, A2, A3, ..., A11, [#permalink]

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05 Mar 2016, 13:08

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In the finite sequence of positive integers A1, A2, A3, ..., A11, each term after the second is the product of the two terms preceding it. If A4 = 18, what is the value of A11?

In the finite sequence of positive integers A1, A2, A3, ..., A11, each term after the second is the product of the two terms preceding it. If A4 = 18, what is the value of A11?

(1) A2 = 3 (2) A6 = 1,944

HI

INFO

each term after the second is the product of the two terms preceding it.

INFERENCE:-

all terms can be brought down to equation of \(A_1\) and \(A_2\).. \(A_3=A_2*A_1\).. \(A_4=A_3*A_2= A_2*A_2*A_1\)... So if we know any TWO terms of the sequence, all numbers can be found

lets see the statements:-

(1) A2 = 3 \(A_4\) is already give.. With\(A_2\), it is SUFF

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

In the finite sequence of positive integers A1, A2, A3, ..., A11, each term after the second is the product of the two terms preceding it. If A4 = 18, what is the value of A11?

(1) A2 = 3 (2) A6 = 1,944

In the original condition, A3 is variable. Since multiplication of the previous 2 terms defines the next term, A4=18. If you figure out A3, you can figure out the rest, which makes 1 variable. In order to match with the number of equations, you need 1 equation. For 1) 1 equation, for 2) 1 equation, which is likely to make D the answer. For 1), A3=6 is derived from (A2)(A3)=18, which is unique and sufficient. For 2), A5 is derived from (A4)(A5)=1,944, which is unique and sufficient. Therefore, the answer is D.

For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
_________________

Re: In the finite sequence of positive integers A1, A2, A3, ..., A11, [#permalink]

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09 Mar 2016, 04:32

Excellent Question.. I made a silly algebraic mistake Here is my approach now A4=A3*A2 => statement 1 gives us A2=> we have A3 too hence we can get every number Similarly as per statement 2 we have A6 => A5*A4 => we have A5 now. So its sufficient too hence D is sufficient

This caution is really good though.
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Re: In the finite sequence of positive integers A1, A2, A3, ..., A11, [#permalink]

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17 Mar 2017, 09:19

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Re: In the finite sequence of positive integers A1, A2, A3, ..., A11, [#permalink]

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03 Aug 2017, 08:22

RITU700 wrote:

In the finite sequence of positive integers A1, A2, A3, ..., A11, each term after the second is the product of the two terms preceding it. If A4 = 18, what is the value of A11?

(1) A2 = 3 (2) A6 = 1,944

Given : finite sequence of positive integers A1, A2, A3, ..., A11, and An = An-1 * An-2 for n >=3 A4 =18 DS : A11

Statement 1 : A2 = 3 A4 = A2*A3 18 = 3*A3 A3 = 6

A1 = A3/A2 = 6/3 = 2 So, sequence is 2,3,6, 18 ........, A11. We can easily calculate A11. SUFFICIENT

Statement 2: A6 = 1944 A5 = A4.A3 = 18.A3 A6= A5. A4 = A5.18 \(1944 = A3. {18}^2\) A3 = 6 So, sequence is A1, A2, 6, 18, 18*6 , 1944,.... A11. We can easily calculate A11. SUFFICIENT