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For a finite sequence of non zero numbers, the number of

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Re: For a finite sequence of non zero numbers, the number of [#permalink] New post 15 Dec 2014, 06:35
Expert's post
Yela wrote:
Hi Brunel,

Quick question. In the sequence below, would consecutive terms not be: 1, 2, -3, 4, 5, -6?
Normally consecutive terms would be: 1,2,3,4,5,... etc, but here we are given negatives:(. This threw off the definition slightly.

Therefore the number of consecutive -ve pairs would be:

2* (-3) = -ve
-4*5 = -ve
5* (-6) = -ve

I get the same answer as you do, but I am just wondering if my approach is correct. As well, if 1, 2, -3, 4, 5, -6 is GMAT's correct definition of consecutive integers, please let me know.

Thanks,
Regards,
Yela

Bunuel 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?

A. 1
B. 2
C. 3
D. 4
E. 5

The questions basically asks: how many pairs of consecutive terms are there in the sequence such that the product of these consecutive terms is negative.

1*(-3)=-3=negative;
-3*2=-6=negative;
5*(-4)=-20=negative.

So there are 3 pairs of consecutive terms.

Answer: C.

Hope it's clear.


Please check this: for-a-finite-sequence-of-non-zero-numbers-the-number-of-97390.html#p750762
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Re: For a finite sequence of non zero numbers, the number of   [#permalink] 15 Dec 2014, 06:35

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