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In a certain sequence, each term, starting with the 3rd term

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In a certain sequence, each term, starting with the 3rd term [#permalink] New post 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 [#permalink] New post 27 Mar 2013, 22:40
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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^4-1) = 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] New post 27 Jun 2013, 22:18
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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] New post 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 [#permalink] New post 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
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Re: In a certain sequence, each term, starting with the 3rd term [#permalink] New post 18 Feb 2014, 21:13
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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] New post 18 May 2014, 09:39
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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
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1/64...8...1/8...1...1/8....1/8...1/64...

Diff b/w 1/64-1/8=-7/64
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Re: In a certain sequence, each term, starting with the 3rd term [#permalink] New post 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 S5-S3?

S5 = (S4)*(S3). Again since S4=1 then S5=S3.

So S5-S3=0

Sufficient

Answer: B

Hope this helps!
Cheers
J :)
Re: In a certain sequence, each term, starting with the 3rd term   [#permalink] 27 May 2014, 05:33
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