SOLUTION: How many GP are possible contaning 27,8 and 12 as three of their terms 1(1) 2(2) 3(4) 4(infintely many)

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Question 1076734: How many GP are possible contaning 27,8 and 12 as three of their terms
1(1)
2(2)
3(4)
4(infintely many)
Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
Infinitely many. Common ratio could be 3%2F2. The term 2%2A2%2A2=8 could be anywhere in the sequence.