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by arocks » Thu Oct 18, 2007 3:26 am
For a finite sequence of non-zero 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

Pls explain. Thanks.

OA - C
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by samirpandeyit62 » Thu Oct 18, 2007 6:25 am
As per my understanding of the question,

nos of changes in sign i.e

1,-3,2,5,-4,-6

1, -3 ---- 1 change

-3, 2 ----- 1 change

5,-4 ----- 1 chnage

= 3 changes

ATQ this should be equal to "the number of pairs of consecutive terms of the sequence for which the product of the two consecutive terms is negative"

i.e. 1,-3 & -3,2 & 5,-4

this is implicit coz if product of two nos is -ve then one has to be -ve & the other +ve i.e a sign change

C
Regards
Samir
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by dingo001 » Thu Oct 18, 2007 7:36 am
1*-3 = -3
-3*2 = -6
2*5 = 10
5*-4=-20
-4*-6=24

above shows 3 results with negative signs. Thus C
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