How many ordered triples $(a, b, c)$ of integers satisfy $1 \leq a<b<c \leq 10$ and $b-a=c-b$ ?
If we let $d=b-a$, then we have $b=a+d$, and since $c-b=b-a=d$, we have $c=b+d=a+2 d$. Thus, $(a, b, c)=(a, a+d, a+2 d)$.

Since $b>a, d>0$, and so we can enumerate the triples by considering possible values of $d$ starting at $d=1$.

When $d=1$, we get the triples $(1,2,3),(2,3,4),(3,4,5),(4,5,6),(5,6,7),(6,7,8),(7,8,9)$, and $(8,9,10)$ for a total of 8 triples.

When $d=2$, we get the triples $(1,3,5),(2,4,6),(3,5,7),(4,6,8),(5,7,9)$, and $(6,8,10)$ for a total of 6 triples.

When $d=3$, we get the triples $(1,4,7),(2,5,8),(3,6,9)$, and $(4,7,10)$ for a total of 4 triples. When $d=4$, we get the triples $(1,5,9)$ and $(2,6,10)$ for a total of 2 triples.

If $d \geq 5$, then since $a$ is at least $1, c=a+2 d \geq 1+2 \times 5=11$. This means $d$ cannot be any larger than 4 , so we have found all triples.

In total, there are $8+6+4+2=20$ triples.

ANSWER: 20