ANSWER
[tex]S_{\infty}=6[/tex]
EXPLANATION
Given:
1. First term (a) = 3
2. Common ration (r) = 3/6
Desired Outcome:
Infinite sum of the geometric sequence.
The formula to calculate the infinite sum of the geometric sequence
[tex]S_{\infty}=\frac{a(1-r^n)}{1-r}[/tex]
Now, as n approaches infinity,
[tex]1-r^n\text{ approaches 1}[/tex]
So,
[tex]\frac{a(1-r^n)}{1-r}\text{ approaches }\frac{a}{1-r}[/tex]
Therefore,
[tex]S_{\infty}=\frac{a}{1-r}[/tex]
Substitute the values
[tex]\begin{gathered} S_{\infty}=\frac{3}{1-\frac{3}{6}} \\ S_{\infty}=\frac{3}{1-\frac{1}{2}} \\ S_{\infty}=\frac{3}{\frac{1}{2}} \\ S_{\infty}=6 \end{gathered}[/tex]
Hence, the infinite sum of the geometric sequence is 6.
