Why is the arithmetic mean of arithmetic sum same as arithmetic mean of the first and the last term of arithmetic sum?

Question

This is used in measuring the average velocity of uniform acceleration. I tried to figure it out myself but i couldn't do it.

Answer 1

You can see this by a direct calculation $$\frac{1}{n} \sum_{i=0}^{n-1} (a + i b) = \frac{1}{n} \left( n a + b \sum_{i=0}^{n-1} i \right) = \frac{1}{n} \left( n a + b \frac{n(n-1)}{2} \right) = a + b \frac{n-1}{2}.$$

$$ \frac{1}{2} (a + a + b (n-1)) = a + b \frac{n-1}{2}. $$

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But probably it is better to regroup à la Gauss $$ \frac{1}{n} \sum_{i=0}^{n-1} (a + i b) = \frac{1}{n} (a + a + b + a + 2 b + \dots + a + b(n-1))=\\ = \frac{1}{n} ((a + a + b (n-1)) + (a + b + a + b(n-2)) + (a + 2 b + a + b(n-3)) + \dots )=\\= \frac{1}{n} \frac{n}{2} (2 a + b (n-1)) = \frac{1}{2} (2 a + b (n-1)) = a + b \frac{n-1}{2}. $$ This is better, in particular because in the first version I have taken for granted the formula $$ \sum_{i=0}^{n-1} i = \frac{n(n-1)}{2} \qquad\text{or}\qquad \frac{1}{n} \sum_{i=0}^{n-1} i = \frac{n-1}{2}, $$ which is the particular case of the formula we are trying to prove for $a = 0$, $b = 1$.