Let the terms be a, ar, ar², ...
The sum to infinity being 27 means that:
[tex]\frac{a}{1-r}=27 \implies a=27(1-r)[/tex]
The sum of the first two terms being 15 means:
[tex]a+ar=15 \\ \\ (27-27r)+r(27-27r)=15 \\ \\ 27-27r+27r-27r^2 =15 \\ \\ 27r^2 -12=0 \\ \\ 27r^2 =12 \\ \\ r^2=\frac{12}{27}[/tex]
Since all the terms are positive, r>0.
[tex]\boxed{r=\frac{2}{3}} \\ \\ \implies a=27\left(1-\frac{2}{3} \right)=\boxed{9}[/tex]
