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?

  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.

answer: 486in^2

step-by-step explanation: