To determine which statements are true, let's consider the sum of n consecutive integers equal to 0.
Let's assume the first integer in the sequence is x. Then the sum of n consecutive integers can be expressed as:
x + (x+1) + (x+2) + ... + (x+n-1) = 0
We can simplify this equation by rearranging and combining like terms:
nx + 1 + 2 + ... + (n-1) = 0
nx + (1 + 2 + ... + (n-1)) = 0
nx + ((n-1)(n-1+1))/2 = 0
nx + ((n-1)(n))/2 = 0
nx + (n^2 - n)/2 = 0
Now, let's analyze the given statements:
I. n is an even number:
If n is an even number, then we can express it as n = 2k, where k is an integer. Substituting this into the equation above:
x(2k) + ((2k)^2 - 2k)/2 = 0
2kx + (4k^2 - 2k)/2 = 0
2kx + 2k^2 - k = 0
2k(x + k) = k
The left-hand side (LHS) is even, but the right-hand side (RHS) is odd (k is not divisible by 2). Therefore, statement I cannot be true.
II. n is an odd number:
If n is an odd number, then we can express it as n = 2k + 1, where k is an integer. Substituting this into the equation above:
x(2k + 1) + ((2k + 1)^2 - (2k + 1))/2 = 0
2kx + x + (4k^2 + 4k + 1 - 2k - 1)/2 = 0
2kx + x + (4k^2 + 2k)/2 = 0
2kx + x + 2k^2 + k = 0
2k(x + 1) + (k + 2k^2) = 0
2k(x + 1) + k(1 + 2k) = 0
Both the LHS and RHS are even since they involve the multiplication of an even number (2k) with an integer. Therefore, statement II can be true.
III. The average (arithmetic mean) of the n integers is 0:
To find the average of n consecutive integers, we divide the sum by n:
(x + (x+1) + (x+2) + ... + (x+n-1))/n = 0
If we simplify this equation:
nx + (1 + 2 + ... + (n-1)) = 0
nx + ((n-1)(n))/2 = 0
This equation is the same as the one we obtained earlier. Therefore, the average of the n integers will be 0 if the sum is 0. So, statement III can be true.
Based on our analysis, the correct answer is:
E. II and III