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April 24, 2014

April 24, 2014

Posted by **Student** on Sunday, October 14, 2012 at 2:53am.

- precal -
**Reiny**, Sunday, October 14, 2012 at 10:19amso you want

1.65 + 1.65(.8) + 1.65(.8)^2 + .. = 8.2

you could just keep adding until you reach 8.2 but that would be less elegant than using a geometric series

a = 1.65

r = .8

Sum(n) = 8.2

a(1 - r^n)/(1-r) = sum(n)

1.65(1 - .8^n)/(1-.8) = 8.2

(1-.8^n)/.2 = 8.2/1.65

1-.8^n = 8.2/1.65*.2

.8^n = 1 - 8.2/1.65*.2

n log .8 = log [1 - 8.2/1.65*.2]

n = log [1- 8.2/1.65*.2] / log .8 = 22.8

22 drives will not do it, so they will need 23 pile drivings

check:

Sum(22) = 1.65 (1-.8^22)/.2 = 8.189

sum(23) = 1.65(1-.8^23)/.2 = 8.2013

my answer is correct.

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