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Sequence

by gmat740 » Mon Jun 15, 2009 8:24 am
For a finite sequence 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 is in sign for the sequence. 1, -3, 2, 5, -4, -6?
OA-3
Imo-9 pairs
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by abhinav85 » Mon Jun 15, 2009 9:41 am
If you arrange these numbers in sequence without taking into
consideration the postives and negatives.......it will be like this

1, -3, 2, 5, -4, -6? = 1 2 -3 -4 5 -6..........

In this scenario we can get 3 variations........

1. 2 x -3 = -6

2. -4 x 5 = -20

3. 5 x -6 = -30.


i don't know its right but according to me thats the only method.
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by mehravikas » Mon Jun 15, 2009 9:16 pm
I dont know whether my approach is right. I did not rearrange the list:

Going through the list as it is:

Number of consecutive pairs where product of two terms is negative are:

(1, -3), (-3, 2) and (5, -4)

therefore there are 3 pairs.
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Re: Sequence

by Vemuri » Mon Jun 15, 2009 11:21 pm
gmat740 wrote:For a finite sequence 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 is in sign for the sequence. 1, -3, 2, 5, -4, -6?
OA-3
Imo-9 pairs
Hi Karan, How did you get 9 pairs?
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by gmat740 » Tue Jun 16, 2009 6:06 am
Gosh I missed the word Consecutive!!
I hope this does not happen again with me!

Thanks everyone
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by aj5105 » Tue Jun 16, 2009 10:39 pm
Should we not re-arrange the terms here?
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