How many A.P.'s with 10 terms are there whose first term is in the set {1, 2, 3} and whose common difference is in the set {1, 2, 3, 4, 5}?

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Solution

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

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