Tuesday

March 31, 2015

March 31, 2015

Posted by **Miracle** on Tuesday, February 18, 2014 at 6:13am.

- Maths G.P. -
**Steve**, Tuesday, February 18, 2014 at 10:21ama = 5

r = 3

Sn = a(1-r^n)/(1-r) = 5(3^n - 1)/2

So, we want

5(3^n-1)/2 > 10^8

3^n-1 > 4*10^7

3^n > 4*10^7 - 1

n > log(4*10^7 - 1)/log3

Now, using base 10 logs, and ignoring the useless -1,

n > (

n > 7+log4)/log3

n > 15.9

So the first 16 terms will sum to more than 10^8

As a sanity check, 5*3^15 = 7*10^7 so I figure just the last two or three terms will produce the desired amount, and the first 13 terms are just noise.

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