TomB wrote:
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. 1
B. 2
C. 3
D. 4
E. 5
this problem is already posted in the forum. My doubt is every body multiplying the negative number with positive number to find the variations. but the question asked for "number of pairs of consecutive terms of the sequence for which the product of the two consecutive terms is negative." for ex:1,-3 are not consecutive . please explain
You are probably mixing consecutive terms in a sequence and consecutive integers: 1 and -3 are not consecutive integers, but they are consecutive terms in the sequence given. See complete solution below.
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
Given sequence: {1, -3, 2, 5, -4, -6}
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;
2*5=10=positive;5*(-4)=-20=negative;
(-4)*(-6)=24=positive.So there are 3 pairs of consecutive terms of the sequence for which the product is negative.
Answer: C.
Hope it's clear.
However per pair the sign change is then happening only twice.