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The numbers are $10a+b,10b+c,$ and $10c+d$. Note that only $d$ can be zero, the numbers $ab$, $bc$, and $cd$ cannot start with a zero, and $a\le b\le c$. To form the sequence, we need $(10c+d)-(10b+c)=(10b+c)-(10a+b)$. This can be rearranged as $10(c-2b+a)=2c-b-d$. Notice that since the left-hand side is a multiple of $10$, the right-hand side can only be $0$ or $10$. (A value of $-10$ would contradict $a\le b\le c$.) Therefore we have two cases: $a+c-2b=1$ and $a+c-2b=0$. Case 1 If $c=9$, then $b+d=8,2b-a=8$, so $5\le b\le 8$. This gives $2593,4692,6791,8890$. If $c=8$, then $b+d=6,2b-a=7$, so $4\le b\le 6$. This gives $1482,3581,5680$. If $c=7$, then $b+d=4,2b-a=6$, so $b=4$, giving $2470$. There is no solution for $c=6$. Added together, this gives us $8$ answers for Case 1. Case 2 This means that the digits themselves are in an arithmetic sequence. This gives us $9$ answers, $$1234,1357,2345,2468,3456,3579,4567,5678,6789.$$ Adding the two cases together, we find the answer to be $8+9=$ $\boxed{17}$.