Solution:

We know $a \leq b \leq c$ by analyzing $ab < bc < cd.$ Also, $bc$ is the average of $ab$ and $cd$ so $$\dfrac {10a+b+10c+d}2 = 10b+c.$$ This means that $$10(c-2b+a)=-b+2c-d,$$ making the right hand side a multiple of $10.$ Thus, it must be $0$ or $10$ since it is digits that satisfy $a \leq b \leq c .$ Thus, we can case on that value.

Case 1: $$-b+2c-d=10$$$$2b-a-c=1$$

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 $b=5,6,7,8.$ This case has $8$ 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. This case makes $9$ solutions.

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

Thus, the correct answer is D .