How to call the average difference between successive values In statistics when you have dataset of values say: $\{1,2,3,4,5,6,7,8,9,10\},$ the mean of differences between every two consecutive observations is $1$. For the dataset $\{1,2,6\}$, it is $2.5$. 
For $\{1,2,3\}$ it is $1$. For $\{1,2,4\},$ it is $1.5$.
Is there a name for this? Some people suggest "moving average," but I'm not sure.
 A: I'm not sure there is a special name for this. But @RoddyMcPhee has suggested
something reasonable. (And I have another suggestion later.)
You may be
interested to know that there are related functions in R statistical
software. The function diff finds the differences between adjacent
observations. The function mean finds the sample mean.
x = c(1,2,3,4,5,6,7,8,9,10)
d.x = diff(x); d.x
## 1 1 1 1 1 1 1 1 1
mean(d.x)
## 1

Notice that this method makes sense only if the values are sorted in increasing
order.
Using the Comment by @G.Sassatelli, sorting is not necessary, but we do need
to know the maximum and the minimum.
(max(x) - min(x))/(length(x)-1)
## 1

Here is a simulation of an application of your idea. We put 1000 observations
at random into the interval $(0,1).$ Then sort them, and view the result as
sequential arrival times of a process. The 'interarrival times' (times
between successive arrivals) are very nearly distributed according to
an exponential distribution with rate $\lambda = 1/1000.$ Accordingly,
the mean of the interarrival times should be about $\mu = 1/\lambda = 0.001.$
y = runif(1000)        # generate 1000 obs from UNIF(0,1)
d.y = diff(sort(y))    # 999 interarrival times
mean(d.y)
## 0.0009988142        # aprx mean interarrival time, exact = 0.001

Here is a histogram of the inter-arrival times along with the 
density function of the exponential distribution with rate 1/1000.

Therefore, in a situation like this where the sorted data values are
viewed as interarrival times of a process, the terminology
you may be looking for is mean interarrival time.
Note: I have to admit a small discrepancy in this example. Because
the simulation forced there to be exactly 1000 arrivals in one unit
of time, the (conditional) distribution of interarrival times is not exactly exponential. 
(However, a Kolmogorv-Smirnov goodness-of-fit test fails to detect the
discrepancy, $\text{P-value }\approx 0.8.$)
