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days
start of day
end of day
1
200.000
120.000
2
320.000
192.000
3
392.000
235.200
4
435.200
261.120
5
461.120
276.672
6
476.672
286.003


days
start of day
end of day
7
486.003
291.602
8
491.602
294.961
9
494.961
296.977
10
496.977
298.186
11
498.186
298.912
12
498.912
299.347
13
499.347
299.608
14
499.608
299.765
15
499.765
299.859
16
499.859
299.915
17
499.915
299.949
18
499.949
299.970
19
499.970
299.982
20
499.982
299.989
The amount of drug (in milligrams) in a person’s bloodstream when 60 percent 
remains from the previous day and an additional 200 mg are added each day.
A pharmacist can modify this initial amount on the first day and observe
changes in the limit of this sum to determine that 80 mg is an appropriate daily
dosage to maintain 200 mg in the bloodstream over time, as shown below. 
days
start of day
end of day
1
80.000
48.000
2
128.000
76.800
3
156.800
94.080
4
174.080
104.448
5
184.448
110.669
6
190.669
114.402
7
194.401
116.641
8
196.641
117.984
9
197.984
118.791
10
198.791
119.274
11
199.274
119.565
12
199.565
119.739
13
199.739
119.843
118
SERIES


days
start of day
end of day
14
199.843
119.906
15
199.906
119.944
16
199.944
119.966
17
199.966
119.980
18
199.980
119.988
19
199.988
119.993
20
199.993
119.996
The amount of drug (in milligrams) in a person’s bloodstream when 60 percent 
remains from the previous day and an additional 80 mg are added each day.
This situation is an example of a geometric series, since the amount remaining in
the bloodstream is affected by a constant ratio of 60 percent. The sum can be re-
written as
days since last dosage
1
2
3
4
80 + 80(0.60)
1
+ 80(0.60)
2
+ 80(0.60)
3
+ 80(0.60)
4
+  . . .
The sum, 
s, can be determined by the equation s =
g
1
(1−r
n
)
1−r
, where 
g
1
is the 
initial dosage, 
r is the constant ratio, and n is the number of days the dosage is
taken. Since the number of days that the drug is taken is unknown, pharmacists
need to examine situations in which the drug is taken indefinitely. Therefore the 
sum of an infinite geometric series is 
s =
g
1
1−r
because
lim
n→∞
g
1
(1−r
n
)
1−r
=
g
1
1−r
when
|r| < 1. In this case, the desired sum, s, is 200 mg, r is 60 percent or 0.60,
and
g
1
is unknown. Substituting the values into the equation, you will get 
200 =
g
1
1−0.60
, and a solution of 
g
1
= 80 mg. Thus the doctor needs to make pre-
scriptions of 80 mg each day in order to maintain the desired dosage of 200 mg.
Geometric series are also used to predict the amount of lumber that can be
cut down each year in a forest to ensure that the number of trees remain at a sta-
ble level. Each year, forest rangers plant seeds for new trees to account for those
chopped down and lost to forest fires. Suppose the ranger wants to know what
proportion of trees they can afford to lose or remove each year if they plant 500
new trees and want to consistently maintain 80,000 trees in the forest. After sub-
stituting
s and g
1
in the formula 
s =
g
1
1−r
, the unknown value for 
r is 0.00625. 
This means that the forest ranger wants to maintain 99.375 percent of the trees
each year. However, an interesting phenomenon is to notice that the forest can
recover from a disaster such as a fire in a reasonably short period of time.
Suppose a fire destroys 35 percent of the trees in the forest, leaving 52,000 trees.
If 500 new trees are planted each year, and 0.625 percent of the total number of

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