Prove the sum of the arithmetic means between 2 numbers, equidistant from the beginning & the end is constant?

Fairly simple. You have 2 numbers, a and z, at the beginning and end. The mean of them is (a+z)/2 = m.

a - - - - - - - -m - - - - - - - -z

Now you want to find the mean of two numbers equidistant from them, say d and w.

a - - -d - - - - -m - - - - -w - - - z

d = a+3 and w = z-3. The mean of d & w is

(d+w)/2 = [(a+3) + (z-3)]/2 = (a+z)/2.

As you can see, if they are equidistant, the "distance" cancels out, and you have the same mean each time. You can also kind of see from the picture that it will always balance like a see-saw as long as you come in the same amount on each side.

OK, you have two numbers, A and B. You can place these in one-to-one correspondence to any two points on a line.

Their arithmetic mean is (A + B)/2. We can place this number, let's call it M, in one-to-one correspondence with another point on the line. M will fall between A and B on the line.

Now we have to prove that distance AM equals distance MB.

The easiest way to do this is to arbitrarily set M at the origin (so its coordinate is zero) and adjust the coordinates of A and B to match.

For instance, if A = 4 and B = 10, M = 7. Now we make the adjustment newx = oldx - 7. M's new coordinate is 0. A and B both have new coordinates. But the formula again was:

(A + B)/2 = M. And M = 0 (now)

A + B must, therefore, equal 0.

The only way for this to happen, A must be the addiitive opposite of B; A = -B.

So the distance from the origin (which is M, now) to A must be the same as the distance from the origin to B. OR AM = MB. QED