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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?

User Dragly
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1 Answer

1 vote

Answer:

88

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.

if we have a geometric series with u1 = 2.1 and r= 1.06, what would the least value-example-1
User HaroldFinch
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5.9k points