SOLUTION: Insert 4 geometric means between 1/8 and 64. I've worked on this so far: an = a₁r^n � 1 a₁ = 1/8, a₆ = 64 64 = (1/8)r⁶⁻' r⁵ = 512 r =

Question 925566: Insert 4 geometric means between 1/8 and 64.
I've worked on this so far:
an = a₁r^n � 1
a₁ = 1/8, a₆ = 64
64 = (1/8)r⁶⁻'
r⁵ = 512
r = 4
an = a₁r^n � 1
an = (1/8)(4)4^n � 1
and I don't know if I'm wrong or not, so please let me know if I am.
Answer by srinivas.g(540) (Show Source): You can put this solution on YOUR website!

a1 = 1/8
a6=64

move 8 to th right

here "r" is not equals to 4

r= 3.48

=
=1.5138

=
=5.26

=18.33
so four geometric means = 0.435 , 1.5138,5.26,18.33
Remarks: As per my observation , r may not be 3.48. I hope there is mistake in the problem. it should be 1/16 instead of 1/8
then r = 4
a2 = 1/4
a3 = 1
a4 = 4
a5 =16

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SOLUTION: Insert 4 geometric means between 1/8 and 64. I've worked on this so far: an = a&#8321;r^n � 1 a&#8321; = 1/8, a&#8326; = 64 64 = (1/8)r&#8310;&#8315;' r&#8309; = 512 r =

SOLUTION: Insert 4 geometric means between 1/8 and 64. I've worked on this so far: an = a₁r^n � 1 a₁ = 1/8, a₆ = 64 64 = (1/8)r⁶⁻' r⁵ = 512 r =