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The sequence a1,a2,…,an,… is such that an=(an−1)(an−3) for

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The sequence a1,a2,…,an,… is such that an=(an−1)(an−3) for [#permalink]

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The sequence a1, a2, …, an, … is such that \(a_n=\sqrt{a_{n-1}*a_{n-3}}\) for all integers n≥4. If \(a_4=16\), what is the value of a6?

(1) a1=2
(2) a2=4

Kudos for a correct solution.
[Reveal] Spoiler: OA

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Re: The sequence a1,a2,…,an,… is such that an=(an−1)(an−3) for [#permalink]

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New post 28 Jan 2015, 09:23
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Is the answer C?

Below is my solution

a4 = sq(a3*a1) option 1 gives us a1 value.--(1)
a6 = sq(a5*a3) and a5 =sq(a4*a2)..option 2 gives us a2 value..combining (1)+(2) gives a6 ..
a2 value is given in option2 hence using

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Re: The sequence a1,a2,…,an,… is such that an=(an−1)(an−3) for [#permalink]

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New post 28 Jan 2015, 09:23
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Is the answer C?

Below is my solution

a4 = sq(a3*a1) option 1 gives us a1 value.--(1)
a6 = sq(a5*a3) and a5 =sq(a4*a2)..option 2 gives us a2 value..combining (1)+(2) gives a6 ..
a2 value is given in option2 hence using :)

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Re: The sequence a1,a2,…,an,… is such that an=(an−1)(an−3) for [#permalink]

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New post 28 Jan 2015, 12:16
IMO- Answer should be C. Please post OA.

Bunuel wrote:
The sequence a1, a2, …, an, … is such that \(a_n=\sqrt{a_{n-1}*a_{n-3}}\) for all integers n≥4. If \(a_4=16\), what is the value of a6?

(1) a1=2
(2) a2=4

Kudos for a correct solution.

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Re: The sequence a1,a2,…,an,… is such that an=(an−1)(an−3) for [#permalink]

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New post 28 Jan 2015, 14:52
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Bunuel wrote:
The sequence a1, a2, …, an, … is such that \(a_n=\sqrt{a_{n-1}*a_{n-3}}\) for all integers n≥4. If \(a_4=16\), what is the value of a6?

(1) a1=2
(2) a2=4

Kudos for a correct solution.



Answer is C:

from stem: a4 =16

from 1: a1 = 2, from stem a4 = sqrt(2*a3) ==> 16 = sqrt(2*a3), so a3 = 128,
we just know the sequence as 2,a2,126,15,a5..., we need a5 to find a6, so NSF

from 2: a2 = 4, as a1 is not known, this statement itslef is insufficient NSF

combined 1+2:
a5 = sqrt(a4*a2) = 8,
sequence is like 2,4,128,16,8,..we have all values to find a6 so C

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Re: The sequence a1,a2,…,an,… is such that an=(an−1)(an−3) for [#permalink]

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New post 02 Feb 2015, 03:17
To Find a6 , we need to know the values of a5 & a3

Stmnt 1 - gives value of a3 - Not Suff
Stmtn 2 - gives value of a5 - Not suff

Combined - Suff

Ans - C

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Re: The sequence a1,a2,…,an,… is such that an=(an−1)(an−3) for [#permalink]

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New post 02 Feb 2015, 03:35
Bunuel wrote:
The sequence a1, a2, …, an, … is such that \(a_n=\sqrt{a_{n-1}*a_{n-3}}\) for all integers n≥4. If \(a_4=16\), what is the value of a6?

(1) a1=2
(2) a2=4

Kudos for a correct solution.


VERITAS PREP OFFICIAL SOLUTION:

One key for this problem, which looks worse than it actually is, is to avoid actually solving for the values. We only need to answer the question of whether a certain value can be solved for, not the more difficult question of what value would actually result.

The sequence rule gives that \(a_6=\sqrt{(a_5)(a_3)}\). In order to obtain the value of a6, we will need to obtain the values of both a5 and a3. Standing alone, a4 cannot produce either.

Adding statement (1) to the mix, we have the values of a1 and a4. Using the sequence rule with n=4 gives \(a_4=\sqrt{(a_3)(a_1)}\). Since we have both a4 and a1, this equation will allow us to solve for a3. From there, however, we can go no further. The values of a1,a3, and a4 do not collectively produce any of the other terms, and, in particular, we cannot solve for a5 with this information. Statement (1) alone is insufficient.

Taking statement (2) alone, we have the values of a2 and a4. Using the sequence rule with n = 5 gives \(a_5=\sqrt{(a_4)(a_2)}\). We have both a4 and a2, so we can get a5. Again, though, we reach a dead end. The values of a2,a4, and a5 do not collectively produce any of the other terms, and, in particular, we cannot solve for a3 with this information. Statement (2) alone is insufficient.

Combining the two statements, we are able to obtain all values from a1 through a5 (and beyond). In particular, statement (1) allows us to solve for a3, and statement (2) allows us to solve for a5. Recalling that \(a_6=\sqrt{(a_5)(a_3)}\), we know that we will be able to obtain a6. The two statements taken together are sufficient.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: The sequence a1,a2,…,an,… is such that an=(an−1)(an−3) for [#permalink]

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New post 12 Sep 2017, 14:28
Bunuel,

Is this approach right.

√(ab)=√a√b ,

So a4= √a3*√a1
a5= √a4*√a2
a6= √a5*√a3
Weare given a4=16
a4= √a3*√a1

STMT 1: a1=2 .

16=√a3*√2

so √a3= 16/√2 = 8√2

STM2: a2=4

a5= √a4*√a2

a5=√16*√4

a5= 8

a6= √a5*√a3
From above we can find the value of a6
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Re: The sequence a1,a2,…,an,… is such that an=(an−1)(an−3) for   [#permalink] 12 Sep 2017, 14:28
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