SOLUTION: The sixth term of a geometric series of positive terms is 10 and the sixteenth term is 0.1, what is the eleventh term?

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Question 1078683: The sixth term of a geometric series of positive terms
is 10 and the sixteenth term is 0.1, what is the eleventh term?

Found 2 solutions by akch2002, Edwin McCravy:
Answer by akch2002(12)   (Show Source): You can put this solution on YOUR website!
We have Tn = a*r^(n-1) where Tn is nth tern, a = 1st term and r = common ratio.
Here 10 = a*r^(6-1)= ar^5 ----(1)
Also 0.1 = a*r^(16-1) = a r^15 ----- (2)
(1)/(2) gives r= 0.01^0.1
and, hence, a = 10/(0.01)^0.5
Hence, 11th term = T11 = ar^10= [10/0.01^0.5]*0.01
= 10(0.01)^0.5
= 1

Answer by Edwin McCravy(20056)   (Show Source): You can put this solution on YOUR website!
Here's a different approach:

The sixth term of a geometric series of positive terms
is 10 and the sixteenth term is 0.1, what is the eleventh term?


Then the series is:



Each term is the preceding term multiplied by the 
common ratio r, so



So we have this system of equations:



We solve the first equation for r5:




We substitute for r5 in the second equation:







Multiply both sides by 10



So





But since we are told the terms are all positive,



and the eleventh term is 1.

Edwin


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