How to find average of consecutive integers | SHSAT Math

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Finding Average of Consecutive integers ( Arithmetic Mean) :

If the numbers have common difference, you can just add the first number with the last number and divide by 2 to get the average of the numbers.

Example: Find the Average of the following numbers:
10, 20, 30, 40, 50
Shortcut way: (10+50)/2 =30

also, when the numbers are consecutive, median is also equal to mean.

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[MrChimili]

This can be easily proven.
Assume x is initial number, y is the difference and there are n numbers after the initial number.
For 1, 3, 5, 7, 9, 11
x=1(initial) y=2(difference) and n=5 (numbers excluding initial)
Above sequence can be written as
x, x + 1*y, x + 2*y, x + 3*y, x + 4*y, x + 5*y x + n*y
Adding this, we get x + n*x + y(1+2+3+4+5....n)
==> x + n*x + y*n*(n+1)/2
==> x(n+1) + yn(n+1)/2
There are n+1 numbers in the sequence ( first number and then n numbers after first number)
Average = x(n+1)/(n+1) + ny(n+1)/2(n+1)
==> x + ny/2
==> 2x/2 + ny/2
==> (2x + ny)/2
==> (x + x+ny)/2
Which is nothing but the short cut described in video.