Consider the set of numbers: a, 2a, 3a, ..., na
where a and n are positive integers. Show that the expression for the mean of this set is
a(n+1)/2

Question

Consider the set of numbers: a, 2a, 3a, ..., na
where a and n are positive integers. Show that the expression for the mean of this set is 
a(n+1)/2

zarkon · Answer

use 1+2+...+n=n(n+1)/2

deoxna · Answer

I have to actually show how (all the work). So far I have that the mean is 
(a + 2a + 3a + .... + na)/n  =  a(1+2+3+....+n)/n. 
But I'm not sure what to do from there. We solved this problem in class a week ago, but I can't remember how we did it, and I remember those were the first steps.

anonymous · Answer

from where you are now, use what Zarkon posted.

deoxna · Answer

Is that some kind of theorem, that the mean of a series is n(n+1)/2?

anonymous · Answer

The sum of the first n natural numbers is:$$1+2+3+\cdots+(n-1)+n=\frac{n(n+1)}{2}$$i wouldnt call it a theorem exactly.  its just a better way to sum the first n natural numbers.  a very important formula to know, helps in a lot of places.

deoxna · Answer

Ok, so then why is the sum of the first n natural numbers n(n+1)/2? Like, how do you arrive at this formula?

zarkon · Answer

$$\begin{matrix}S= & 1+2+3+\cdots (n-1)+n\S= & n+(n-1)+\cdots +3+2+1\end{matrix}$$

add together

$$2S=(n+1)+(n+1)+\cdots (n+1)=n(n+1)$$

$$S=\frac{n(n+1)}{2}$$

anonymous · Answer

im sure zarkon is writing out a proof as we speak :)  there are a couple os ways to prove it.

deoxna · Answer

Thanks! It took me a little while to get ho you were getting 2S=n(n+1), but then I realized you were adding the "columns" of the top two equations vertically.