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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]

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21 May 2015, 00:15
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.

Hope it's clear.

We are taking -3*2 but why we are not taking -6*5
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Re: For a finite sequence of non zero numbers, the number of  [#permalink]

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21 May 2015, 01:15
Baten80 wrote:
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.

Hope it's clear.

We are taking -3*2 but why we are not taking -6*5

Hi Baten80,

The question talks about the -ve sign of product of consecutive terms. The sequence is already ordered and hence -6 and 5 are not consecutive terms whereas -3 and 2 are.

Hope it's clear

Regards
Harsh
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Re: For a finite sequence of non zero numbers, the number of  [#permalink]

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21 May 2015, 03:22
Baten80 wrote:
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.

Hope it's clear.

We are taking -3*2 but why we are not taking -6*5

Was answered before: for-a-finite-sequence-of-non-zero-numbers-the-number-of-97390.html#p750762

Hope it helps.
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Joined: 22 Jan 2018
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Re: For a finite sequence of non zero numbers, the number of  [#permalink]

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05 Mar 2019, 04:47
BG wrote:
if we order the sequence in increasing order -6,-4,-3,1,2,5 i see only one pair for which the product of two consecutive numbers is negative
i would go with 1

I also went with this approach and got 1. I thought "sequence" means strictly in "ascending" or in "descending" order.

Is it not the case?
Re: For a finite sequence of non zero numbers, the number of   [#permalink] 05 Mar 2019, 04:47

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

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