How do I solve this Geometric Series problem?
"The second term of a geometric series of positive terms is 100 and the fourth term is 64. Find the number of terms that must be added if the total is to exceed four times the first term."
I really don't understand this. Can someone provide a solution please, with method shown?
Thanks 
Comments
The second term is 10^2 The fourth term is 8^2
So I'm going to guess that the series is 11^2, 10^2, 9^2, 8^2...
So the first term I'm suggesting is 11^2 which is 121.
Four times 121 is 484
1st term + 2nd term 11^2 + 10^2 = 221
total + 3rd term 221 + 9^2 = 302
total + 4th term 302 + 8^2 = 366
total + 5th term 366 + 7^2 = 415
total + 6th term 415 + 6^2 = 451
total + 7th term 451+5^2 = 476
total + 8th term 476 +4^2 = 492
So by a simple iterative technique we have established that we need to add up 8 terms to exceed 4 times the 1st term.
A geometric series progresses like so:
a , ar , ar^2 , ar^3 , ar^4 , .... , ar^n
So, the second term would be a * r, the fourth term would be a * r^3
a * r = 100
a * r^3 = 64
(ar^3) / (ar) = 64 / 100
r^2 = 64/100
r = +/- 8/10
r = +/- 4/5
Let's assume that r is positive
a * r = 100
a * (4/5) = 100
a = 100 * 5 / 4
a = 125
We want to find when 4 * 125 is exceeded by the sum of the geometric terms:
4 * 125 = sum(125 * (4/5)^n , n = 0 , t)
4 * 125 = 125 * sum( (4/5)^n , n = 0 , t)
4 = sum( (4/5)^n , n = 0 , t)
4 = (4/5)^0 + (4/5)^1 + (4/5)^2 + (4/5)^3 + (4/5)^4 + .... (4/5)^t
t = 7
Assume that r = -4/5
4 = sum( (-4/5)^n , n = 0 , t)
4 = 1 - (4/5) + (16/25) - (64/125) + ....
t does not exist.
a_2 = 100 = ar
a_4 = 64 = ar^3
r^2 = 64/100
r = 8/10 = 4/5
a(4/5) = 100
a = 125
4a = 500
S_n = 125 (1 - 0.8)^n)/(1 - 0.8) = 125 (1 - 0.8^n)/0.2 > 500
125 (1 - 0.8^n) > 100
1 - 0.8^n > 0.8
0.2 > 0.8^n
ln 0.2 > n ln 0.8
ln 0.2/ln 0.8 > n
7.21256... > n
n = 7
This doesn't seem right, nor does it check. But I don't know what I did wrong.