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In a certain sequence, each term, starting with the 3rd term
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27 Mar 2013, 21:53
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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
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27 Jun 2013, 22:18
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
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27 Mar 2013, 22:40
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
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04 Feb 2014, 04:04
(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
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18 Feb 2014, 16:29
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
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18 Feb 2014, 21:13
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
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18 May 2014, 09:39
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
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27 May 2014, 05:33
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
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04 Aug 2014, 09:23
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
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29 Dec 2014, 19:09
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
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29 Dec 2014, 19:35
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
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24 Nov 2016, 19:48
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
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24 Nov 2016, 20:56
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
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31 Oct 2017, 10:30
OE
We have been asked for the difference between the 6th and 3rd terms of a sequence. There is little we can do to rephrase the question, but it is important to note that we don’t necessarily need to know the exact values of these terms to find their difference. We can solve as long as we know the relative values of the two terms, i.e. how much larger or smaller the 6th term is than the 3rd.
(1) INSUFFICIENT: We have been given the relationship between the 1st and 2nd terms. If we let x be the value of the 2nd term, then the 1st term is 8x. Since each term is determined by multiplying the previous two terms, we can build the whole sequence in terms of x. We can begin by multiplying 8x, the1st term, and x, the 2nd term, to get the 3rd term, 8x2. If we continue to build, we get the following sequence (starting from the 1st term): 8x, x, 8x2, 8x3, 82x5, 83x8
The difference between the 6th and 3rd terms is 83x8 − 8x2, but we can’t calculate the actual value without knowing the value of x.
(2) SUFFICIENT: At first glance, this statement does not appear to provide much helpful information. However, let’s look at what this means for the rest of the sequence. The 4th term is 1. We can find the 5th term by multiplying the 3rd and 4th terms. Let’s use n to represent the unknown 3rd term, and see what happens. 3rd term: n 4th term: 1 5th term: n*1 = n 6th term: 1*n = n
The presence of the 1 in the 4th spot produces a pair of “copycat” terms afterward! In other words, the 6th and 3rd terms are equal. Therefore, the difference between them is 0 and the correct answer is B.
Alternatively, we could approach this problem using Smart Numbers.
(1) INSUFFICIENT: We can try different numbers for the first two terms to see whether we get different results.
If we make the first two terms 8 and 1, we produce this sequence: 8, 1, 8, 8, 82, 83 What other numbers can we try? If we increase our 2nd term by 1, we get 16 and 2, but from there we will swiftly find ourselves multiplying some very large numbers! Fortunately, we are free to adjust downward and use fractions. Let’s try 4 and 1/2: 4, ½, 2, 1, 2, 2 Since 83 − 8 is clearly not the same value as 2 − 2, statement 1 is not sufficient.
(2) SUFFICIENT: While we can now use a number rather than a variable to represent the 3rd term, the result will look a lot like our previous solution: 3rd term: 4 4th term: 1 5th term: 4*1 = 4 6th term: 1*4 = 4
No matter what number we plug in, the 3rd and 6th terms must be equal, so their difference will always be 0.
The correct answer is B.



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Re: In a certain sequence, each term, starting with the 3rd term
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08 Jan 2018, 06:17
1) a1,a2,a3,a4,a5,a6 8a,a,8a^2,8a^3,64a^5,512a^8 no info about a INSUFF 2) a4=1 a3=a3 a4=1 a5=a3 a6=a3 a6a3 =0 SUFFICIENT B
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