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I run into this problem and have no clue what to do and how [#permalink]
28 Sep 2008, 15:15

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 _________________

The one who flies is worthy. The one who is worthy flies. The one who doesn't fly isn't worthy

Re: Sequence problem - Nightmare [#permalink]
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

Re: Sequence problem - Nightmare [#permalink]
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.

gmatclubot

Re: Sequence problem - Nightmare
[#permalink]
28 Sep 2008, 20:58