How to do this sequences and series question

An arithmetic series has a first term of -4 and a common difference of 1. A geometric series has a first term of 8 and a common ratio of 0.5. After how many terms does the sum of the arithmetic series exceed the sum of the geometric series?

Note first that the infinite geometric series would have a sum of 8/(1 - 0.5) = 16,
and that after the first 4 terms of the sequence, namely 8, 4, 2 and 1, we have
already reached a sum of 15. By the 4th term of the arithmetic sequence the series
sum will still be negative. So what we need to find is the value n such that the
arithmetic series will have reached a sum of 16.

Now for an arithmetic series the sum is S(n) = (n/2)*(2a + (n-1)*d), where in this
case a = -4 and d = 1. So we need to find the smallest n such that

S(n) = (n/2)*(2*(-4) + (n - 1)*1) > 16 ---->

n*(-9 + n) > 32 ---->

n^2 - 9n - 32 > 0 ----> use the quadratic formula to solve for

n^2 - 9n - 32 = 0 ----> n = (9 +/- sqrt(81 + 128))/2, and since we need n > 0

we have n = (9 + sqrt(209))/2 = 11.72. But since we need an integer value

we need to round up to n = 12 to be certain that the A.S. exceeds the G.S..

For n = 12 the A.S. will have a sum of (12/2)*(2*(-4) + (12 - 1)*1) = 6*3 = 18,

while the G.S. will have a sum between 15 and 16.

(Note that for n = 11 the A.S. will have a sum of only (11/2)*(-8 + 10) = 11.)

whoa