Respuesta :
[tex]\begin{gathered} a)S_2=\text{ the sum of the first two terms} \\ S_2=a_1+a_2=15 \\ S_2=a_1(\frac{r^2-1}{r-1}) \\ r=\text{ratio} \\ 15=a_1(\frac{r^2-1}{r-1})\text{ , but} \\ r^2-1=(r+1)(r-1),\text{ hence} \\ 15=a_1(\frac{(r+1)(r-1)}{r-1}) \\ 15=a_1(r+1) \\ \text{Sum to infinity} \\ S_{\infty}=\frac{a_1}{1-r} \\ 27=\frac{a_1}{1-r} \\ 27(1-r)=a_1\text{ (2)} \\ U\sin g\text{ (2) in (1)} \\ 15=27(1-r)(r+1) \\ (1-r)(r+1)=r+1-r^2-r \\ (1-r)(r+1)=1-r^2 \\ 15=27(1-r^2) \\ 15=27-27r^2 \\ 15+27r^2=27 \\ 27r^2=27-15 \\ 27r^2=12 \\ r^2=\frac{12}{27} \\ r^2=\frac{4}{9} \\ \sqrt{r^2}=\sqrt{\frac{4}{9}} \\ r=\frac{2}{3} \\ \text{The value of the common ratio is }\frac{2}{3} \\ b)For\text{ }a_1 \\ 27(1-r)=a_1 \\ a_1=27(1-\frac{2}{3}) \\ a_1=27(\frac{1}{3}) \\ a_1=9 \\ The\text{ value of the first term is 9} \\ \end{gathered}[/tex]
