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10 Nov 2007, 11:42
Kudos
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For a finite series 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
A. One
B. Two
C. Three
D. Four
E. Five
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10 Nov 2007, 11:55
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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.
In English --->
In a set the product of the two consecutive terms is negative (i.e -1*1 = -1 but 1 and -1 are not consecutive or 1*2 = 2 consecutive but this is not negative)
in the list 1,-3,2,5,-4,-6 you have
2*-3 = -6
-4*5 = -20
5*-6 = -30
-3*4 = -12
the answer is (D)
In English --->
In a set the product of the two consecutive terms is negative (i.e -1*1 = -1 but 1 and -1 are not consecutive or 1*2 = 2 consecutive but this is not negative)
in the list 1,-3,2,5,-4,-6 you have
2*-3 = -6
-4*5 = -20
5*-6 = -30
-3*4 = -12
the answer is (D)
10 Nov 2007, 12:13
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KillerSquirrel
KillerSquirrel,
I like the way you translated the question. The correct answer is C. You got it all right until you multiplied -3*4. There is no positive 4 in the sequence. So, the answer is 3 variations.
Thanks for the explanation
