RA911 wrote:
Let's say that the 6 consecutive integers are x, x + 1, x + 2, x + 3, x + 4, and x + 5. We are told that their product is equal to Q. So,
(x) * (x + 1) * (x + 2) * (x + 3) * (x + 4) * (x + 5) = Q
We can check each option individually to find out which ones could be true:
Option A):
The product of these 6 consecutive integers can be greater than 0 when all of the integers are also greater than 0. If x > 0, then:
(x) * (x + 1) * (x + 2) * (x + 3) * (x + 4) * (x + 5) = Q > 0
Therefore, Option A) could be true.
Option B):
Similarly, the product of these 6 consecutive integers can be equal to 0 when any one of the consecutive integers is equal to 0. If x = 0, then:
(x) * (x + 1) * (x + 2) * (x + 3) * (x + 4) * (x + 5) = Q = 0
Therefore, Option B) could be true.
Option C):
When some of the consecutive integers are less than 0, then the product of them can not be negative. For example, if x = -3, then the consecutive integers will be -3, -2, -1, 0, 1, and 2. Their product will be:
(x) * (x + 1) * (x + 2) * (x + 3) * (x + 4) * (x + 5) = (-3) * (-2) * (-1) * (0) * (1) * (2)
(x) * (x + 1) * (x + 2) * (x + 3) * (x + 4) * (x + 5) = 0
That's why, the Option C) can not be true.
Option D):
The mean of the set of these 6 consecutive integers x, x + 1, x + 2, x + 3, x + 4, and x + 5 will be:
Mean = (x)+(x+1)+(x+2)+(x+3)+(x+4)+(x+5)6
Mean = (x+x+x+x+x+x)+(1+2+3+4+5)6
Mean = (6x)+(15)6
Mean = 3∗(2x+5)2∗3
Mean = (2x+5)2
The median of the set of these 6 consecutive integers x, x + 1, x + 2, x + 3, x + 4, and x + 5 will be:
Median = (x+2)+(x+3)2
Median = (2x+5)2
Mean = Median
It means that the Option D) is true as well.
Option E):
Finally, we can see that the mean will be less than 0 when most of the consecutive integers are also less than 0. For example, if x = -3, then:
Mean} = (2∗(−3)+5)2
Mean = (−6+5)2
Mean = (−1)2
That's why, the Option E) can not be true.
Hence, the correct answers are Option A), Option B), and Option D).
Hi, cannot option E be true ? for example 1,2,3,4,5,6 have a mean of 3.5 which is positive.