The first term and the common ratio of a geometric series are denoted, respectively, by a and r where a , $r \in \mathbb{Q}$. Given that the third term is 9 and the sum to infinity is 64, find the value of a and the value of r .

we are given that $a{r^2} = 9$ and $\frac{a}{{1 - r}} = 64$     A1

dividing, ${r^2}(1 - r) = \frac{9}{{64}}$     M1

$64{r^3} - 64{r^2} + 9 = 0$     A1

$r = 0.75,{\text{ }}a = 16$     A1A1

[5 marks]