From $ab<bc<cd$, we have $a\le b\le c.$ The arithmetic-sequence condition is $$\begin{aligned} 2(10b+c)&=(10a+b) \\ &\quad +(10c+d), \end{aligned}$$ which rearranges to $$10(a-2b+c)=-b+2c-d.$$ The right side is a multiple of $10.$ Because $b\le c$ and all four symbols are digits with $b\ge1,$ it lies between $-9$ and $17$. Hence it is either $0$ or $10.$

Case 1: $$\begin{aligned} -b+2c-d&=10, \\ a&=2b-c+1. \end{aligned}$$

We can look at the possible values of $c.$

$c=6:$ $b+d=2, 2b-5=a.$ Thus, $b\leq 2$ from the first equation, but can't work for the second equation.

$c=7:$ $b+d=4, 2b-6=a.$ Thus, $b\leq 4$ from the first equation, and $b > 3$ from the second equation. This makes one case for $b=4.$

$c=8:$ $b+d=6, 2b-7=a.$ Thus, $b\leq 6$ from the first equation, and $b > 3$ from the second equation. This makes three cases for $b=4,5,6.$

$c=9:$ $b+d=8, 2b-8=a.$ Thus, $b\leq 8$ from the first equation, and $b > 4$ from the second equation. This makes four cases for $b=5,6,7,8.$ Altogether this case gives $$\begin{gathered} 2470,1482,3581,5680, \\ 2593,4692,6791,8890, \end{gathered}$$ for $8$ solutions.

Case 2: $$-b+2c-d=0,$$$$2b-a-c=0,$$ which means the digits are an arithmetic sequence.

If the difference is $1,$ then $1 \leq a \leq 6$ makes $6$ solutions.

If the difference is $2,$ then $1 \leq a \leq 3$ makes $3$ solutions. A difference of at least $3$ would force $d=a+3r>9.$ This case therefore gives $9$ solutions.

In total, the number of solutions is $8+9 = 17.$

Thus, the correct answer is D .