<s>[INST] 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.
 [/INST] 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
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