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02 Apr 2007, 01:41
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For a finite sequence of non zero numbers, the number of variations in signs is defined as the number of pairs of consecutive terms for which the product of the two consecutive terms is negative. What is the number of varations for the number in sequence 1,-3, 2,5,-4,-6?
a)1
b)2
c)3
d)4
e)5
a)1
b)2
c)3
d)4
e)5
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what is this question asking? 
06 Apr 2007, 16:32
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the question asks if you multiply each term in the series with the next term, how many such products will be negative ?
so let us multipy the first term byu the second term: 1 * -3 = -3 (negative)
2nd term * 3rd term = -3 * 2 = -6 (negative)
3rd term * 4th term = 2*5 = 10 (positive)
4th term * 5th term = 5* -4 = -20 (negative)
5th term * 6th term = -4*-6 = 24 (positive)
total = 3 negatives. therefore, C
so let us multipy the first term byu the second term: 1 * -3 = -3 (negative)
2nd term * 3rd term = -3 * 2 = -6 (negative)
3rd term * 4th term = 2*5 = 10 (positive)
4th term * 5th term = 5* -4 = -20 (negative)
5th term * 6th term = -4*-6 = 24 (positive)
total = 3 negatives. therefore, C
