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# I run into this problem and have no clue what to do and how

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Senior Manager
Joined: 04 Aug 2008
Posts: 369
I run into this problem and have no clue what to do and how [#permalink]

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28 Sep 2008, 16:15
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

I run into this problem and have no clue what to do and how to approach it (if you answer please explain the steps)

For a finite sequence of nonzero numbers, the number of variations in sign is defined as the number of pairs of consecutive terms of the sequence for which the product of the two consecutive terms is negative. What is the number of variations in sign for the sequence 1, -3, 2, 5, -4, -6 ?

one
two
three
four
five
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The one who flies is worthy. The one who is worthy flies. The one who doesn't fly isn't worthy

Manager
Joined: 22 Sep 2008
Posts: 119
Re: Sequence problem - Nightmare [#permalink]

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28 Sep 2008, 17:43

i am not sure but i solved by taking two random series.

1:1 ,-2 (total 1 negative product 1*-2 & 1 sign change)

2: 1,3,-2 (total 1 negative product ,3*-2 & 1 sign change)

Senior Manager
Joined: 04 Aug 2008
Posts: 369
Re: Sequence problem - Nightmare [#permalink]

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28 Sep 2008, 18:14
sorry buddy thats not the answer
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The one who flies is worthy. The one who is worthy flies. The one who doesn't fly isn't worthy

Intern
Joined: 26 May 2008
Posts: 8
Location: Chicago, IL
Re: Sequence problem - Nightmare [#permalink]

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28 Sep 2008, 18:23
Intern
Joined: 24 Aug 2008
Posts: 8
Re: Sequence problem - Nightmare [#permalink]

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28 Sep 2008, 18:32

In the sequence 1, -3, 2, 5, -4, -6

This sign changes 3 times ( therefore product of 2 adjacent numbers is negative )

1 to -3
-3 to 2
5 to -4

What's OA ?
Intern
Joined: 17 Aug 2008
Posts: 12
Re: Sequence problem - Nightmare [#permalink]

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28 Sep 2008, 18:36
For a finite sequence of nonzero numbers, the number of variations in sign is defined as the number of pairs of consecutive terms of the sequence for which the product of the two consecutive terms is negative. What is the number of variations in sign for the sequence 1, -3, 2, 5, -4, -6 ?

one
two
three
four
five

(1*-3)=-3 negative
(-3*2)=-6 negative
(2*5)=10 positive
(5*-4)=-20 negative
(-4*-6)=24 positive
3 Negative products of consecutive terms
Intern
Joined: 03 Mar 2008
Posts: 44
Re: Sequence problem - Nightmare [#permalink]

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28 Sep 2008, 18:37
IMO three
OA ?

if correct could expand on my method
Current Student
Joined: 28 Dec 2004
Posts: 3314
Location: New York City
Schools: Wharton'11 HBS'12
Re: Sequence problem - Nightmare [#permalink]

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28 Sep 2008, 19:10
i am going with 3..

the number of sign changes..lets see...1, -3, 2, 5, -4, -6

-1*3=-3 -4*2=-8 -6*2=-12
Senior Manager
Joined: 04 Aug 2008
Posts: 369
Re: Sequence problem - Nightmare [#permalink]

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28 Sep 2008, 19:16
OA is three

I also guessed this answer but cannot explain it. Seems like bertlacy comes w logical explanation
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The one who flies is worthy. The one who is worthy flies. The one who doesn't fly isn't worthy

Manager
Joined: 22 Sep 2008
Posts: 119
Re: Sequence problem - Nightmare [#permalink]

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28 Sep 2008, 21:47
oops.. big blunder by me.. i have not see the series and just start calculating
VP
Joined: 05 Jul 2008
Posts: 1377
Re: Sequence problem - Nightmare [#permalink]

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28 Sep 2008, 21:58
bertlacy wrote:
For a finite sequence of nonzero numbers, the number of variations in sign is defined as the number of pairs of consecutive terms of the sequence for which the product of the two consecutive terms is negative. What is the number of variations in sign for the sequence 1, -3, 2, 5, -4, -6 ?

one
two
three
four
five

(1*-3)=-3 negative
(-3*2)=-6 negative
(2*5)=10 positive
(5*-4)=-20 negative
(-4*-6)=24 positive
3 Negative products of consecutive terms

Agree that 3 is correct.
Re: Sequence problem - Nightmare   [#permalink] 28 Sep 2008, 21:58
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# I run into this problem and have no clue what to do and how

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