For a finite sequence of non zero numbers, the number of variations in the 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?
1)1
2)2
3)3
4)4
5)5
3
Why is (1, -3) considerd a set
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For a finite sequence of non zero numbers
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Anurag@Gurome
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venmic wrote:For a finite sequence of non zero numbers, the number of variations in the 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?
1)1
2)2
3)3
4)4
5)5
3
Why is (1, -3) considerd a set
We get a negative integer when one term is positive and the other is negative.
1 * -3 = -3
-3 * 2 = -6
5 * -4 = -20
Therefore, required number of variations = 3
The correct answer is [spoiler](3)[/spoiler].
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Thank you very much for the solution. I had the same problem and still do. Could someone please explain why 1 and -3 are consecutive terms while 5 and -6 are not? Actually how do we identify the pair of consecutive terms.
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tomada
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In a sequence of 'N' numbers, the set of consecutive terms = (1st term and 2nd term), (2nd term and 3rd term), ......, ( Nth-1 term, Nth term)
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