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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.

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