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Finite sequence

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by j_shreyans » Thu Oct 16, 2014 8:19 pm
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 term is negative. What is the number of variation in sign for the sequence 1,-3,2,5,-4,-6

A)One
B)Two
C)Three
D)Four
E)Five

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by Brent@GMATPrepNow » Thu Oct 16, 2014 8:50 pm
j_shreyans 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 term is negative. What is the number of variation in sign for the sequence 1,-3,2,5,-4,-6

A)One
B)Two
C)Three
D)Four
E)Five

OAC
We're asked to look at every pair of CONSECUTIVE numbers. If the product of that pair is NEGATIVE, this counts as one variation.

Let's examine the pairs of consecutive numbers:

1 and -3: product is negative
-3 and 2: product is negative
2 and 5: product is positive
5 and -4: product is negative
-4 and -6: product is positive

Since 3 pairs have negative products, the correct answer is C

Cheers,
Brent
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by Mathsbuddy » Fri Oct 17, 2014 1:19 am
Between consecutive terms, I have placed the operator sign in brackets to indicate a positive or negative product (NB: the "product" = the answer to multiplied values):

1 (-) -3 (-) 2 (+) 5 (-) -4 (+) -6

There are 3 negatives, so answer = C
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by GMATGuruNY » Fri Oct 17, 2014 1:29 am
j_shreyans 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 term is negative. What is the number of variation in sign for the sequence 1,-3,2,5,-4,-6

A)One
B)Two
C)Three
D)Four
E)Five
{-1, 3, 2, 5, -4, -6}
The product of two consecutive terms will be negative if the two terms have DIFFERENT SIGNS.
In the sequence above, the following pairs of consecutive terms have different signs:
1, -3
-3, 2
5, -4.
Thus, the number of variations = 3.

The correct answer is C.
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by Jeff@TargetTestPrep » Wed Dec 13, 2017 9:29 am
j_shreyans 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 term is negative. What is the number of variation in sign for the sequence 1,-3,2,5,-4,-6

A)One
B)Two
C)Three
D)Four
E)Five
We are given the following sequence of numbers: 1, -3, 2, 5, -4, -6.

Every time a pair of consecutive terms product a negative product we have a "variation in sign". We must determine how many variations in sign are in the sequence.

1 x (-3) = -3, so this is a variation in sign

(-3) x 2 = -6, so this is a variation in sign

5 x (-4) = -20, so this is a variation in sign

Thus, there is a total of 3 variations in sign.

Answer: C

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