Steadily accumulating number formula This is quite likely a super simple question, but I lack the descriptive power to be able to find the answer on Google!
Consider the following two number sequences:
00, 10, 20, 30, 40, 50, 60, 70, 80, 90
45, 45, 45, 45, 45, 45, 45, 45, 45, 45

Both sequences have a total value of 450, but one accumulates steadily in multiples of 10, whilst the second remains constant at 45.
The second is easy to ascertain, divide the total value by the number of instances in the list. But the first sequence I'm stuck on - can anyone help me understand the formula for ascertaining the accumulation value (in this instance 10) from 0?
My goal is to be able to take any number and divide it x number of times, with the first number in the sequence being 0 and steadily increasing - apportioning greater and greater values to those later in the sequence.

It's an arithmetic sequence (i. e. every next element is created from the previous with adding the same number)  with the well-known formula for its  sum
$$a_1 + a_2 + \cdots + a_n = {n \over 2}(a_1+a_n)$$
So, as $a_1= 0,\ n = 10,\ $ and $\ a_{10} = 90$ you will obtain
$$00 + 10 +20 + \cdots + 90 = {10\over2}(0 + 90)= 5\cdot 90 = 450$$

Addendum:
If you know the first element ($a_1$), number of elements ($n$) and their sum ($s$), you may obtain the last element ($a_n$) by substituting the known values into the formula, and then isolate $a_n$:
$$a_n = {2s\over n}-a_1$$
You may then compute the difference $d$ between adjacent members as
$$d = {a_n-a_1\over n-1}$$
to consecutively obtain members between $a_1$ and $a_n$: 
\begin{aligned}
a_2 &= a_1+d\\a_3 &= a_2+d\\
&\cdots
\end{aligned}
(To the question in your comment.)
