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In a certain sequence, each term, starting with the 3rd term [#permalink]
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27 Mar 2013, 22:53
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In a certain sequence, each term, starting with the 3rd term, is found by multiplying the previous two terms. What is the difference between the 6th and 3rd terms in the sequence? (1) The 1st term is equal to 8 times the 2nd term. (2) The 4th term is equal to 1.
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Re: In a certain sequence, each term, starting with the 3rd term [#permalink]
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mun23 wrote: In a certain sequence, each term, starting with the 3rd term, is found by multiplying the previous two terms. What is the difference between the 6th and 3rd terms in the sequence?
(1) The 1st term is equal to 8 times the 2nd term.
(2) The 4th term is equal to 1.
Need explanation From F.S 1, consider this series : 8, 1, 8, 8, 64, 64*8. Again, consider the series 2, 1/4, 1/2, 1/8, 1/16, 1/(8*16). Clearly, the difference of the 6th and the 3rd term is different for them. Insufficient. From F.S 2, let the series be \(a,b,ab,ab^2,a^2b^3,a^3b^5\). Now we know that \(ab^2 = 1\). The required difference =\(a^3b^5  ab = ab(a^2b^41) = ab[(ab^2)^2 1]\)= 0.Sufficient. B.
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Re: In a certain sequence, each term, starting with the 3rd term [#permalink]
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27 Jun 2013, 23:18
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mun23 wrote: In a certain sequence, each term, starting with the 3rd term, is found by multiplying the previous two terms. What is the difference between the 6th and 3rd terms in the sequence?
(1) The 1st term is equal to 8 times the 2nd term. (2) The 4th term is equal to 1. The question is: 1st term = \(x\) 2nd term = \(y\) 3rd term = \(xy\) 4th term = \((xy)*y = xy^2\) 5th term = \((xy^2)*(xy) = x^2y^3\) 6th term = \((x^2y^3)*(xy^2) = x^3y^5\) Have equation: 6th term  3rd term \(= x^3y^5  xy = xy(x^2y^4 1)\) Statement 1: The 1st term is equal to 8 times the 2nd term ==> \(x = 8y\) Replace back to equation: 6th term  3rd term \(= 8yy((8y)^2y^4 1) = 8y^2(64y^6  1)\) ==> NOT Sufficient because we don't know y Statement 2: The 4th term is equal to 1 ==> \(=xy^2 = 1\). Replace back to equation: 6th term  3rd term \(= xy((xy^2)^2  1) = xy(1  1) = 0\) ==> Sufficient Hence, B is correct. Hope it helps.
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Re: In a certain sequence, each term, starting with the 3rd term [#permalink]
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04 Feb 2014, 05:04
(1) First term is 8 times the second term. could be 8 and 1 or 64 and 8 or 2 and 1/4, which would all lead to different sequences, varying the differences between the 6th and the 3rd term.. IS.
(2) Since the 4th term is 1 and has to be a product of the 3rd and 2nd term, all terms are 1, so the difference will be zero. SUFF.
Hence answer B.
I guess it was a lucky pick for me, but I didn't took long time setting up equations and stuff. Went straight as I described it, took me about 40 seconds!



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Re: In a certain sequence, each term, starting with the 3rd term [#permalink]
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18 Feb 2014, 17:29
unceldolan wrote: (1) First term is 8 times the second term. could be 8 and 1 or 64 and 8 or 2 and 1/4, which would all lead to different sequences, varying the differences between the 6th and the 3rd term.. IS.
(2) Since the 4th term is 1 and has to be a product of the 3rd and 2nd term, all terms are 1, so the difference will be zero. SUFF.
Hence answer B.
I guess it was a lucky pick for me, but I didn't took long time setting up equations and stuff. Went straight as I described it, took me about 40 seconds! Why should all terms be 1? Do we have to assume that all terms are integers here? What if S2 = 1/3 and S3= 3 then S4 = (1/3)(3) = 1 Can anyone explain second statement? Thanks a lot Cheers J



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Re: In a certain sequence, each term, starting with the 3rd term [#permalink]
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18 Feb 2014, 22:13
jlgdr wrote: unceldolan wrote: (1) First term is 8 times the second term. could be 8 and 1 or 64 and 8 or 2 and 1/4, which would all lead to different sequences, varying the differences between the 6th and the 3rd term.. IS.
(2) Since the 4th term is 1 and has to be a product of the 3rd and 2nd term, all terms are 1, so the difference will be zero. SUFF.
Hence answer B.
I guess it was a lucky pick for me, but I didn't took long time setting up equations and stuff. Went straight as I described it, took me about 40 seconds! Why should all terms be 1? Do we have to assume that all terms are integers here? What if S2 = 1/3 and S3= 3 then S4 = (1/3)(3) = 1 Can anyone explain second statement? Thanks a lot Cheers J No, you don't have to assume anything. What you say is absolutely possible though it doesn't change our answer. "S2 = 1/3 and S3= 3 then S4 = (1/3)(3) = 1" The sequence could be: 9, 1/3, 3, 1, 3, 3, .... But in any case \(t_6  t_3 = 0\) Note why: Statement 2 tells us that \(t_4 = 1\) \(t_1, t_2, t_3, 1,\) what will be \(t_5\)? It will be \(t_3 * 1 = t_3\) What will be \(t_6\)? It will be \(1*t_5 = 1*t_3 = t_3\) So in any case the sixth term will be same as the third term. Once you have a 1 in the sequence, all following terms will be equal to the term just preceding 1.
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Re: In a certain sequence, each term, starting with the 3rd term [#permalink]
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18 May 2014, 10:39
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The answer is B...
First Term to Sixth 1..1..1...1...1...1
Diff b/w 6th and 3rd is 1 (1)=0 _______________________ 1/64...8...1/8...1...1/8....1/8...1/64...
Diff b/w 1/641/8=7/64 



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Re: In a certain sequence, each term, starting with the 3rd term [#permalink]
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27 May 2014, 06:33
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Shouldn't be hard at all, let's see So we need to find S6  S3 Now, S6 = (S5)(S4) So (S5)(S4)  S3? Statement 1 Clearly insufficient. We don't have much information here as given that S=8(S2) Statement 2 Now, here's where it gets interesting. We have a specific value S4=1. Now, since S4=1 we are being asked what is S5S3? S5 = (S4)*(S3). Again since S4=1 then S5=S3. So S5S3=0 Sufficient Answer: B Hope this helps! Cheers J



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Re: In a certain sequence, each term, starting with the 3rd term [#permalink]
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04 Aug 2014, 10:23
mun23 wrote: In a certain sequence, each term, starting with the 3rd term, is found by multiplying the previous two terms. What is the difference between the 6th and 3rd terms in the sequence?
(1) The 1st term is equal to 8 times the 2nd term. (2) The 4th term is equal to 1. Statement 1 is clearly insufficient as it tells nothing about 4th or 5th term Statement 2 says that the 4th term is 1 Let the 3rd term be anything.... lets say the 3rd term is 'a' and the 4th term is 1 Thus 5th term would be a*1.. which is equal to a And 6th term would be again 5th term*4th term that is a*1=a Thus no matter what the 3rd term be.... if the 4th term is 1 then the 3rd and 6th term would always be equal.. thus the difference would be 0  Sufficient.
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Re: In a certain sequence, each term, starting with the 3rd term [#permalink]
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29 Dec 2014, 20:09
Could someone explain why the 3rd term is not 1 when the 4th is 1? Doesn't the 4th term have to be the product of the 2nd and the 3rd term?



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Re: In a certain sequence, each term, starting with the 3rd term [#permalink]
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29 Dec 2014, 20:35
dina98 wrote: Could someone explain why the 3rd term is not 1 when the 4th is 1? Doesn't the 4th term have to be the product of the 2nd and the 3rd term? \(T_4 = T_3 * T_2 = 1\) \(T_3\) could be anything and \(T_2\) would take a corresponding value. For example, \(T_3\) could be 2 and \(T_2\) would be 1/2. Their product would be 1. Their product needs to be 1, they both independently don't need to be 1.
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Re: In a certain sequence, each term, starting with the 3rd term [#permalink]
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24 Nov 2016, 20:48
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pqhai wrote: mun23 wrote: In a certain sequence, each term, starting with the 3rd term, is found by multiplying the previous two terms. What is the difference between the 6th and 3rd terms in the sequence?
(1) The 1st term is equal to 8 times the 2nd term. (2) The 4th term is equal to 1. The question is: 1st term = \(x\) 2nd term = \(y\) 3rd term = \(xy\) 4th term = \((xy)*y = xy^2\) 5th term = \((xy^2)*(xy) = x^2y^3\) 6th term = \((x^2y^3)*(xy^2) = x^3y^5\) Have equation: 6th term  3rd term \(= x^3y^5  xy = xy(x^2y^4 1)\) Statement 1: The 1st term is equal to 8 times the 2nd term ==> \(x = 8y\) Replace back to equation: 6th term  3rd term \(= 8yy((8y)^2y^4 1) = 8y^2(64y^6  1)\) ==> NOT Sufficient because we don't know y Statement 2: The 4th term is equal to 1 ==> \(=xy^2 = 1\). Replace back to equation: 6th term  3rd term \(= xy((xy^2)^2  1) = xy(1  1) = 0\) ==> Sufficient Hence, B is correct. Hope it helps. Thanks, I did it exactly the same way . It encourages me to know that I think in the proper direction under stressed situations.



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In a certain sequence, each term, starting with the 3rd term [#permalink]
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24 Nov 2016, 21:56
From (1): a1= 8x, a2=x => a3=8x^2 and a6=8^3*x^8 => not sufficient From (2): a1 a2 a3 a4(=1) => a5=a3 and a6=1*a5=a3 => the difference between a3 and a6 = 0 ==> B is the answer



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Re: In a certain sequence, each term, starting with the 3rd term [#permalink]
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04 Jan 2017, 15:49
jlgdr wrote: Shouldn't be hard at all, let's see So we need to find S6  S3 Now, S6 = (S5)(S4) So (S5)(S4)  S3? Statement 1 Clearly insufficient. We don't have much information here as given that S=8(S2) Statement 2 Now, here's where it gets interesting. We have a specific value S4=1. Now, since S4=1 we are being asked what is S5S3? S5 = (S4)*(S3). Again since S4=1 then S5=S3. So S5S3=0 Sufficient Answer: B Hope this helps! Cheers J Best answer hands down...




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