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The number of variations in sign...PS

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by adwaitkasbekar » Tue Dec 21, 2010 11:01 pm
Can anyone please explain how this problem is to be solved?

For a finite set 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 consecutive terms is negative. What is the number of variations in sign for the sequence 1, -3, 2, 5, -4, -6?
a. 1
b. 2
c. 3
d. 4
e. 5
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by Night reader » Tue Dec 21, 2010 11:27 pm
adwaitkasbekar wrote:Can anyone please explain how this problem is to be solved?

For a finite set 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 consecutive terms is negative. What is the number of variations in sign for the sequence 1, -3, 2, 5, -4, -6?
a. 1
b. 2
c. 3
d. 4
e. 5
if I am not mistaken this problem is mixed symbolism and number properties (+ve/-ve)*(+ve/-ve)

if we consider (1;-3) (-3;2) (2;5) (5;-4) (-4;-6) then we get in total 3 (-ve)

IOM C
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by adwaitkasbekar » Tue Dec 21, 2010 11:30 pm
the answer is indeed C.

I got confused by the language.

Thanks for helping me out.
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by N:Dure » Tue Dec 21, 2010 11:31 pm
adwaitkasbekar wrote:
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 consecutive terms is negative. What is the number of variations in sign for the sequence 1, -3, 2, 5, -4, -6?
a. 1
b. 2
c. 3
d. 4
e. 5
I think the bolded part is what he's looking for. So pairs of consecutive terms in the sequence whose product is negative will be (1, -3) (-3, 2) (5,-4)

Number of variations = 3
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by NathaliaC » Mon Jul 02, 2012 9:13 am
This language is indeed very confusing. In my case, while solving I rewrote the sequence in order because the problem mentions "sequence of consecutive terms", so I rewrote to -6, -4, -3, 1, 2, 5. This led to only 1 product of consecutive terms that is negative (-3 * 1). Wrong answer.
-Nathália
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