need help solving geometric series questions 
A geometric sequence has its first term as $10000$ and a fourth term as $−7290$.
  If the pattern continues forever, find the sum of the terms in the sequence. 

I know that the $n^{th}$ term is found by $$t_n=a_1r^{n-1}$$ where $a_1$ is the first term and $r$ is the common ratio. 
Thus the fourth term is given by $$10000 * r^3 = -7290$$
so $$r^3 = -0.729$$ $$r= -0.9$$
So the sum of the series is: $$S_ \infty = \frac{10000}{1-(-0.9)}$$ 
Is this correct?


*The sum of the first 2018 terms of a geometric sequence is 200. The sum of
the first 4036 terms is 380. Find the sum of the first 6054 terms

 A: For #1 Yes. You correctly used the formula for the sum of an infinite geometric series. It works because $|-0.9|<1.$
For #2, let's try to write it out. 
The first 2018 terms would be $a+ar^1+ar^2...ar^{2017}$
The first 4036 terms would be $a+ar^1...ar^{2017}...ar^{2018}+ar^{2019}..ar^{4035}$
Notice that this is simply $(a+ar^1+ar^2...ar^{2017})(1+r^{2018})$.
You can figure out that $1+r^{2018}=1.9$ because the first $2018$ terms add up to $200$ and the first $4036$ to $380$.
Therefore, $r^{2018} = 0.9$.
Finally, the first 6054 terms, as you may be able to tell, is equal to $(a+ar^1+ar^2...ar^{2017})(1+r^{2018}+r^{4036})=200(1+0.9+0.9^2)=200(2.71)=\boxed{542}$
A: For your second question, let $S_n$ be sum of first $nth$ terms and r be the common ratio, then we have
$ S_{2018} = 200$ and $ S_{4036} = 380$
Note that 
$$ S_{4036} = S_{2018} + r^{2018} * S_{2018}$$
$$ 380 = 200 + r^{2018} * 200$$
$$ r^{2018} = 0.9$$
Also note that
$$ S_{6054} = S_{2018} + r^{2018} * S_{2018} + r^{4036} * S_{2018} $$
$$ S_{6054} = 200+ 0.9 * 200+ 0.9^2 * 200 $$
$$ S_{6054} = 542$$
