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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-> 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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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
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
