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For a finite sequence of nonzero numbers, the number of

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Manager
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For a finite sequence of nonzero numbers, the number of [#permalink] New post 03 Jan 2009, 21:12
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 ?

(A) One
(B) Two
(C) Three
(D) Four
(E) Five


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Manager
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Re: Consecutive Numbers [#permalink] New post 04 Jan 2009, 13:36
it is, can you explain how you arrived at that solution?

thanks
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Re: Consecutive Numbers [#permalink] New post 04 Jan 2009, 15:49
you add the count.. if the product of the 2 consecutive numbers = negative.
here are the possibilities:
1st term x 2nd term = negative... that's 1
2nd term x 3rd term = negative... that's 2
3rd x 4th = positive
4th x 5th = negative..that's 3
5th x 6th = positive

so your answer is 3.
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Re: Consecutive Numbers [#permalink] New post 05 Jan 2009, 02:11
answer is C
multiply the consecutive numbers and count when the sign of the multiplied number is negative.
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Re: Consecutive Numbers [#permalink] New post 05 Jan 2009, 10:48
thanks for your detailed explanation cul3s.

When I first encountered this question I re-organized the numbers in order of magnitude, when I should of just used them the way they were given. I got misled since it says "consecutive"

thanks again.
Re: Consecutive Numbers   [#permalink] 05 Jan 2009, 10:48
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For a finite sequence of nonzero numbers, the number of

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