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For a certain sequence, tn=-1(tn-1+1)^-1 for all values of n

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For a certain sequence, tn=-1(tn-1+1)^-1 for all values of n [#permalink] New post 02 Nov 2005, 18:01
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For a certain sequence, tn=-1(tn-1+1)^-1 for all values of n such that n => 2. Given that t1=1, how much greater is the product of the first 200 values in this sequence than the sum of the first 200 values in this sequence?

1
68
98
105
228


Give it a :shot
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Last edited by Titleist on 02 Nov 2005, 22:03, edited 1 time in total.
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 [#permalink] New post 02 Nov 2005, 21:56
I am not able to get any headway n finding t2.... :)

where is our frens lexieqv,qsr,himalaya,gmatt73..???
any hint titleist?
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Re: SEetheconseQUENCE PS [#permalink] New post 02 Nov 2005, 21:58
Titleist wrote:
For a certain sequence, tn=-1(tn-1+1)^-1 for all values of n such that n >= 2 :!: Given that t1=1, how much greater is the product of the first 200 values in this sequence than the sum of the first 200 values in this sequence?

1
68
98
105
228


Give it a :shot

from the formula we can calculate
t2= -1/2 , t3=-2, t4= 1 , t5=-1/2 , t6=-2, t7= 1, this repeats over and over with interval of 3
the sum is t2+ t3+ t4+ ... + t200 = ( t2+t3+t4) + ( t5+t6+t7) + ....+ (t197+t198+t199)+t200
Each sum in the ( ) has the value of -1/2+(-2)+1= -3/2
We have 66 groups in all --> the sum except for t200 is -3/2*66= -99
---> the whole sum is -99+t200= -99-1/2= -99.5

The product of t2*t3*t4= -1/2*-2* 1= 1
We have 66 such groups in all --> the product of these 66 groups is 1
--> the product of these 66 groups with t200= 1*(-1/2) = -1/2
---> the product- the sum= -1/2 -(-99.5) = 99

The slight difference is because I assumed that n>=2 ...coz if we don't have n=2, we can't never calculate t2 and thus t3,t4,t5 ......

Last edited by laxieqv on 02 Nov 2005, 22:00, edited 1 time in total.
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 [#permalink] New post 02 Nov 2005, 21:59
duttsit wrote:
I am not able to get any headway n finding t2.... :)

where is our frens lexieqv,qsr,himalaya,gmatt73..???
any hint titleist?


yeah where is everyone?

as for the hint. find the pattern for the first 10 terms. you'll be able to find a pattern once you sum vs. mulitplying.
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 [#permalink] New post 02 Nov 2005, 22:01
hey, are you sure n > 2 ... I think it should be n > =2 :!:
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 [#permalink] New post 02 Nov 2005, 22:02
laxieqv wrote:
hey, are you sure n > 2 ... I think it should be n > =2 :!:


good find yes - let me make the edit. :beat
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 [#permalink] New post 02 Nov 2005, 22:03
well, please see my solution which is right above your last :wink:
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 [#permalink] New post 02 Nov 2005, 22:08
Titleist wrote:
laxieqv wrote:
hey, are you sure n > 2 ... I think it should be n > =2 :!:


good find yes - let me make the edit. :beat


hey..you made me stare at this problem from last 1 hour... to find elusive t2 while only thing we can know is t1 and t3,t4,...

be back from the break :)
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 [#permalink] New post 02 Nov 2005, 22:18
laxieqv wrote:
well, please see my solution which is right above your last :wink:


Very close Laxi, you probably guessed 98.

here is why it's 98 vs. 99

t1 = 1
t2 = -1/2
t3 = -2

t4 = 1
t5 = -1/2
t6 = -2
t10 = 1
..so forth

The sequence repeats itself every 3 terms.

The sum of of the sequence for the first 3 terms is -1.5 and it is -1.5 for every subsequent 3 terms.

The product of the three terms is 1 and repeats itself every three terms.

If we take 200 terms and divided by the number of cycles, 3, then we get 66. However 66 x 3 is 198 thus we have to add or multiply the 199th and 200th terms.

The product
1^66 = 1 x 1 (199th term) x -0.5 (200th term) = -0.5 (is the product of the first 200 terms of the sequence

The sum
-1.5 x 66 = -99 + 1 (199 term) + -0.5 (200th term) = -98.5

The product - The sum = -0.5 - (-98.5) = 98
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http://www.nytimes.com/2005/11/21/inter ... prexy.html

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 [#permalink] New post 02 Nov 2005, 22:19
duttsit wrote:
Titleist wrote:
laxieqv wrote:
hey, are you sure n > 2 ... I think it should be n > =2 :!:


good find yes - let me make the edit. :beat


hey..you made me stare at this problem from last 1 hour... to find elusive t2 while only thing we can know is t1 and t3,t4,...

be back from the break :)


SORRRY!!!! :oops:
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http://www.nytimes.com/2005/11/21/inter ... prexy.html

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 [#permalink] New post 02 Nov 2005, 22:41
Titleist wrote:
laxieqv wrote:
well, please see my solution which is right above your last :wink:


If we take 200 terms and divided by the number of cycles, 3, then we get 66. However 66 x 3 is 198 thus we have to add or multiply the 199th and 200th terms.



Hey ,aren't we considering n from 2 to 200???? there are 200-2+1= 199 numbers in the set! and thus 66 groups and only t200 left!
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 [#permalink] New post 02 Nov 2005, 22:58
laxieqv wrote:
Titleist wrote:
laxieqv wrote:
well, please see my solution which is right above your last :wink:


If we take 200 terms and divided by the number of cycles, 3, then we get 66. However 66 x 3 is 198 thus we have to add or multiply the 199th and 200th terms.



Hey, aren't we considering n from 2 to 200???? there are 200-2+1= 199 numbers in the set! and thus 66 groups and only t200 left!


No - I think the question explicitly asks for the first 200 values not just 199 values, so you would have to start from T1. The question tells you that n=>2 because they are giving you a starting point from which you can derive the sequence from T1. OA from Manhattan Gmat stands at 98. I double checked the question.
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 [#permalink] New post 02 Nov 2005, 23:03
Titleist wrote:
laxieqv wrote:
Titleist wrote:
laxieqv wrote:
well, please see my solution which is right above your last :wink:


If we take 200 terms and divided by the number of cycles, 3, then we get 66. However 66 x 3 is 198 thus we have to add or multiply the 199th and 200th terms.



Hey, aren't we considering n from 2 to 200???? there are 200-2+1= 199 numbers in the set! and thus 66 groups and only t200 left!


No - I think the question explicitly asks for the first 200 values not just 199 values, so you would have to start from T1. The question tells you that n=>2 because they are giving you a starting point from which you can derive the sequence from T1. OA from Manhattan Gmat stands at 98. I double checked the question.


ok, i accept that ...anyway this one is not tough but tricky :-D ...and it's EXTREMELY tricker as you omitted n=2 :twisted:
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 [#permalink] New post 02 Nov 2005, 23:04
laxieqv wrote:
Titleist wrote:
laxieqv wrote:
Titleist wrote:
laxieqv wrote:
well, please see my solution which is right above your last :wink:


If we take 200 terms and divided by the number of cycles, 3, then we get 66. However 66 x 3 is 198 thus we have to add or multiply the 199th and 200th terms.



Hey, aren't we considering n from 2 to 200???? there are 200-2+1= 199 numbers in the set! and thus 66 groups and only t200 left!


No - I think the question explicitly asks for the first 200 values not just 199 values, so you would have to start from T1. The question tells you that n=>2 because they are giving you a starting point from which you can derive the sequence from T1. OA from Manhattan Gmat stands at 98. I double checked the question.


ok, i accept that ...anyway this one is not tough but tricky :-D ...and it's EXTREMELY tricker as you omitted n=2 :twisted:


Ok Ok! I'm sorry about that - I assure you that there was no malice involved - it was just an innocent mistake!

:knight
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http://www.nytimes.com/2005/11/21/inter ... prexy.html

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 [#permalink] New post 02 Nov 2005, 23:08
Laxi, try the Honey Bee probability question - you'll enjoy that one. :kisswink
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 [#permalink] New post 02 Nov 2005, 23:09
oh my ....your that emoticon tells thousands of words ...btw, it's great fun to have you playing some jokes around :wink: ..But for this thread, I stop chit-chatting, if not I'm gonna get "fired" by CEO 8-)
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 [#permalink] New post 02 Nov 2005, 23:11
laxieqv wrote:
oh my ....your that emoticon tells thousands of words ...btw, it's great fun to have you playing some jokes around :wink: ..But for this thread, I stop chit-chatting, if not I'm gonna get "fired" by CEO 8-)


Fired? He'd better give you a nice severance package Laxieqv.
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"Wow! Brazil is big." —George W. Bush, after being shown a map of Brazil by Brazilian president Luiz Inacio Lula da Silva, Brasilia, Brazil, Nov. 6, 2005

http://www.nytimes.com/2005/11/21/inter ... prexy.html

  [#permalink] 02 Nov 2005, 23:11
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