if we have a geometric series with u1 = 2.1 and r= 1.06, what would the least value of n be such that Sn > 5543?
1 answer:
Answer:
88
Step-by-step explanation:
Write the expression for the sum in the relation you want.
Sn = u1(r^n -1)/(r -1) = 2.1(1.06^n -1)/(1.06 -1)
Sn = (2.1/0.06)(1.06^n -1) = 35(1.06^n -1)
The relation we want is ...
Sn > 5543
35(1.06^n -1) > 5543 . . . . substitute for Sn
1.06^n -1 > 5543/35 . . . . divide by 35
1.06^n > 5578/35 . . . . . . add 1
n·log(1.06) > log(5578/35) . . . take the log
n > 87.03 . . . . . . . . . . . . . . divide by the coefficient of n
The least value of n such that Sn > 5543 is 88.
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