a = 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.
Find the sum of the nth terms of a G.P 5+15+45+...........What is the smallest number of the terms which will give a total of more than 10^8.
1 answer