Let form an infinite geometric progression such that all of its terms are positive integers, and the product of the first 4 terms is 64. Find .
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Let the common ratio of the GP be r , then we have:
a 1 ⋅ a 2 ⋅ a 3 ⋅ a 4 a 1 ⋅ a 1 r ⋅ a 1 r 2 ⋅ a 1 r 3 a 1 4 r 6 = 6 4 = 6 4 = 2 6
For a 1 , a 2 , a 3 and a 4 to be integers, both a 1 and r must also be integers. For a 1 4 r 6 = 2 6 , the solution must be a 1 = 1 and r = 2 . Therefore, a 1 = 1 , a 2 = 2 , a 3 = 4 , a 4 = 8 and a 5 = 1 6 .