geometric progression Help me to solve this question :

The sum of first two terms of a positive geometric progression is $9$ times the sum of the following two terms. Given that the sum of the first four terms is $\frac{320}{9}$.
Find
  
  
*
  
*the first term and the common ratio
  
*the sum of fifth term to infinity 
  

I know that:
$$a+ar=9(ar^2+ar^3)\tag 1$$
And:
$$a+ar+ar^2+ar^3=\frac{320}{9}\tag 2$$
But I am unsure where to continue from here.
 A: We know that $a + ar = 9(ar^2 + ar^3)$ and $a + ar + ar^2 + ar^3 = 320/9$. However the second equation can be rewritten as $10(ar^2 + ar^3) = 320/9$ so that $ar^2 + ar^3 = 32/9$. Dividing by $a$ in the first equation gives us $1 + r = 9r^2 + 9r^3 \implies (9r^2 - 1)(1+r) = 0 \implies r = \pm \frac{1}{3} , -1$ but $-1$ gives you an alternating sequence. Can you figure out the rest from here?
A: You're almost there. First, the common ratio $r$ must be positive (since all terms are positive), that is, $\color\red{r>0}$. Start with equation $(1)$ by dividing by $a$ to get $$1+r=9(r^2+r^3).$$ Next, apply factoring to get $$1+r=9r^2(1+r).$$ Then cancel $1+r$ (since $r>0$ and so $1+r\neq 0$) to get $$1=9r^2$$ so that $9r^2-1=0$ and since $r>0$, so $r\neq -\frac{1}{3}$ and hence $$r=\frac{1}{3}.$$ Equation $(2)$ is equivalent to $$\frac{a(1-r^4)}{1-r}=\frac{320}{9}.$$ Hence,
$$a=\frac{320}{9}\cdot\frac{1-r}{1-r^4}=\frac{320}{9}\cdot\frac{2/3}{80/81}=\frac{320}{9}\cdot\frac{27}{40}=24.$$ This answer part 1.
For part 2, we have
$$\begin{align}
24(1/3)^4+24(1/3)^5+24(1/3)^6+\dots=\frac{24(1/3)^4}{1-\frac{1}{3}}=\frac{4}{9}.
\end{align}$$
