Three different numbers are chosen from 1, 2, 3, …, 20. In how many ways can they be chosen so that they form an arithmetic progression (in some order)?

The middle number is the average of the other two, so the first and last must have the same parity. Same-parity pairs: ${}^{10}{C}_2+{}^{10}{C}_2=45+45=90$; each gives exactly one middle term. (180 counts each triple in both directions.)