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

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I run into this problem and have no clue what to do and how [#permalink] New post 28 Sep 2008, 15:15
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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Re: Sequence problem - Nightmare [#permalink] New post 28 Sep 2008, 16:43
is answer 1 ?

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)

Please confirm it..
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Re: Sequence problem - Nightmare [#permalink] New post 28 Sep 2008, 17:14
sorry buddy thats not the answer
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Re: Sequence problem - Nightmare [#permalink] New post 28 Sep 2008, 17:23
is the answer three?
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Re: Sequence problem - Nightmare [#permalink] New post 28 Sep 2008, 17:32
Answer must be 3.

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 ?
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Re: Sequence problem - Nightmare [#permalink] New post 28 Sep 2008, 17:36
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

The answer is three:
(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
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Re: Sequence problem - Nightmare [#permalink] New post 28 Sep 2008, 17:37
IMO three
OA ?

if correct could expand on my method
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Re: Sequence problem - Nightmare [#permalink] New post 28 Sep 2008, 18:10
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
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Re: Sequence problem - Nightmare [#permalink] New post 28 Sep 2008, 18:16
OA is three

I also guessed this answer but cannot explain it. Seems like bertlacy comes w logical explanation
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Re: Sequence problem - Nightmare [#permalink] New post 28 Sep 2008, 20:47
oops.. big blunder by me.. i have not see the series and just start calculating :(
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Re: Sequence problem - Nightmare [#permalink] New post 28 Sep 2008, 20:58
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

The answer is three:
(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, 20:58
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I run into this problem and have no clue what to do and how

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