Respuesta :
We are given the following infinite geometric series
[tex]6+3+\frac{3}{2}+\frac{3}{4}+\cdots[/tex]The general form of an infinite geometric series is given by
[tex]a_1+a_1r+a_1r^2+a_1r^3+\cdots[/tex]Where a1 is the first term and r is the common ratio.
The first term is equal to 6
The common ratio is the division of any two consecutive numbers in the infinite geometric series.
[tex]\frac{3}{6}=\frac{1}{2}[/tex]You can also take any other two consecutive numbers in the series and you will get the same common ratio.
[tex]\frac{\frac{3}{2}}{3}=\frac{3}{2}\cdot\frac{1}{3}=\frac{3}{6}=\frac{1}{2}[/tex]As expected, we still got the same common ratio as before.
How do I find out whether an infinite geometric series converges?
An infinite geometric series converges if the absolute value of the common ratio is less than 1.
[tex]\begin{gathered} |r|<1 \\ \frac{1}{2}<1 \\ 0.5<1 \end{gathered}[/tex]Since the absolute value of the common ratio is less than 1, the infinite geometric series converges.
How do I find its sum?
The sum of this infinite geometric series is given by
[tex]S=\frac{a_1}{1-r}[/tex]Substitute the values of first term a1 and common ratio r.
[tex]S=\frac{6}{1-\frac{1}{2}}=\frac{6}{\frac{1}{2}}=6\cdot\frac{2}{1}=12[/tex]Therefore, the sum of this infinite geometric series is 12.
