SOLUTION: Find the 10th term of a geometric sequence whose 5th term is 81 and whose ninth term is 16. I solved it as 16 = 81r⁵⁻¹ 16 = 81r⁴ r = -2/3, 2/3 I used the positi

Question 1188249: Find the 10th term of a geometric sequence whose 5th term is 81 and whose
ninth term is 16.
I solved it as
16 = 81r⁵⁻¹
16 = 81r⁴
r = -2/3, 2/3
I used the positive value and did
a₁₀= 81(2/3)⁶⁻¹
a₁₀= 32/3
Is this correct? Thank you
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
The first term is (1.5)^4*81=410.0625 or 410(1/16)
The 10th term is 32/3, as you found.
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It works for r=-(2/3), only the 10th term is -32/3. a1 is the same.

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