ANSWER
D.
[tex]720[/tex]
EXPLANATION
The infinite geometric series given to us is,
[tex]\sum_{n=1}^{\infty}180\times (\frac{3}{4})^{n-1}[/tex]
The first term of this series is,
[tex]a_1=180\times (\frac{3}{4})^{1-1}[/tex]
This implies that,
[tex]a_1=180\times (\frac{3}{4})^{0}[/tex]
[tex]a_1=180[/tex]
[tex]a_2=180\times (\frac{3}{4})^{2-1}[/tex]
[tex]a_2=180\times (\frac{3}{4})^{1}[/tex]
[tex]a_2=135[/tex]
The common ratio of this sequence,
[tex]r = \frac{a_2}{a_1}[/tex]
[tex]r = \frac{135}{180} [/tex]
[tex]r = \frac{3}{4} [/tex]
The sum to infinity of this series is given by the formula,
[tex]S_{\infty}=\frac{a_1}{1 - r}[/tex]
We substitute the above values to get,
[tex]S_{\infty}=\frac{180}{1 - \frac{3}{4} }[/tex]
This simplifies to
[tex]S_{\infty}=\frac{180}{ \frac{1}{4} }[/tex]
This implies that,
[tex]S_{\infty}=180 \times 4[/tex]
[tex]S_{\infty}=720[/tex]
The correct answer is D.