How many A.P.'s with 10 terms are there whose first term is in the set {1, 2, 3}

Number of ways of selecting the first term from the set {1, 2, 3} = 3

Corresponding to each of the selected first terms, the number of ways of selecting the common difference  from the set {1, 2, 3, 4, 5} = 5

$\therefore$ Total number of AP's that can be formed $=3 \times 5=15$