SOLUTION: the sixth term of a geometric series of positive number is 10 and the sixteenth term is 0.1. What is the eleventh term? Find its sum to infinity.

Question 984035: the sixth term of a geometric series of positive number is 10 and the sixteenth term is 0.1. What is the eleventh term? Find its sum to infinity.
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
a sub n=a1 * r^n-1
a6=10
a16=0.1
Let a6 be the first term of the modified sequence.
0.1=10*r^10
10^-2=r^10=0.01
r= 0.01^(1/10)=0.6309. Without rounding.
a1=100
a11=1
Sum=a1 (1-r^n)/1-r
=100*(.9)/1-r
=90/(1-r); 1-r=0.3691(rounded)
unrounded, answer is 243.87.
If one keeps the same ratio, which isn't quite 0.6, this works out exactly. If one rounds to 0.6, it does not.

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