How do you find the geometric means in the sequence #32, , , , , 1#?

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Answer 1 · 2017-03-09T21:25:13

#a_n = 6^(2/5 (6 - n))#

If #36# is a term of a geometric sequence, then

#36 = c cdot r#

but we know also

#1=c cdot r^6#

dividing term to term we obtain

#36=r^-6# so #r=1/root(6)(36) = 1/6^(2/5)# and

#c = 36/r = 36 cdot 6^(2/5)#

and the sequence is

#a_n = 6^(2/5 (6 - n))#

How do you find the geometric means in the sequence #32, __, __, __, __, 1#?