Answered

There are four geometric means between 3 and 729. Find the fourth term of the geometric sequence.

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AnonymousBetaviz

a_n = (a_1)r^(n-1)

a_6 = (a_1)r^(6-1)

729 = 3r^5

r^5 = 243

r= 3

a_4 = (a_1)(3)^(4-1)

a_4 = 3•3^3

=3•27

a_4 = 81