A interesting combinatorics problem...

How many arithmetic progressions with 10 10 terms are there whose first term is in the set { 1 , 2 , 3 } \{1,2,3\} and whose common difference is in the set { 1 , 2 , 3 , 4 , 5 } \{1,2,3,4,5\} ?


The answer is 15.

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1 solution

Culver Kwan
Oct 11, 2020

If you fix the first term and common difference of a A.P. , you get the whole sequence. so there are 3 × 5 = 15 3\times5=\boxed{15} such A.P. .

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