3 digit numbers whose one digit is the average of the other two 
Question: How many three-digit numbers are composed of three distinct digits such that one digit is the average of the other two?

How can this be solved with an arithmetic sequence?
I can see that any two digits must be same parity to produce the even sum. Also any two of the digits must be divisible by $2$. if $\overline{abc}$ is the digit, $a+b=2c$ and doing that for all three digits gives three equations. Thats as far as I can get.

Can be done by simple trial and error
(Note that plus sign is only used to distinguish the triplets of digits that can be used to form the desired number)
$0+2,4,6,8+1,2,3,4$
$1+3,5,7,9+2,3,4,5$
$2+4,6,8+3,4,5$
$3+5,7,9+4,5,6$
$4+6,8+5,6$
$5+7,9+6,7$
$6+8+7$
$7+9+8$
In total we get 20 possible triplets of digits that can be permuted in $3!=6$ different ways over 3 possible places. This gives us $20×6=120$
However, the numbers formed from the 4 triplets with $0$, ie, $0+2,4,6,8+1,2,3,4$, will have $2/6$ permutations in which the unit digit is the one $=0$. But this can't be since the numbers is a 3 digit number.
Hence we will have to subtract $4×2=8$ from $120$ to give our final answer of $112$ which matches the data provided in the other answer
