Answer:
The geometric series is [tex]8,\frac{8}{3} ,\frac{8}{9},\frac{8}{27},\frac{8}{81}[/tex]
The sum of the geometric series is 12.06
Step-by-step explanation:
First term of geometric series = 8
Common Ratio = 1/3
The formula used to find next term is: [tex]a_n=a_1r^{n-1}[/tex]
Our series has five terms, so we need to find 2nd, 3rd, 4th and 5th term
2nd term is:
[tex]a_n=a_1r^{n-1}\\a_2=8(\frac{1}{3})^{2-1}\\a_2=8(\frac{1}{3})^{1}\\a_2=\frac{8}{3}[/tex]
3rd term is:
[tex]a_n=a_1r^{n-1}\\a_3=8(\frac{1}{3})^{3-1}\\a_3=8(\frac{1}{3})^{2}\\a_3=\frac{8}{9}[/tex]
4th term is:
[tex]a_n=a_1r^{n-1}\\a_4=8(\frac{1}{3})^{4-1}\\a_4=8(\frac{1}{3})^{3}\\a_4=\frac{8}{27}[/tex]
5th term is:
[tex]a_n=a_1r^{n-1}\\a_5=8(\frac{1}{3})^{5-1}\\a_5=8(\frac{1}{3})^{4}\\a_2=\frac{8}{81}[/tex]
So, The geometric series is [tex]8,\frac{8}{3} ,\frac{8}{9},\frac{8}{27},\frac{8}{81}[/tex]
Now, Finding the sum of geometric series
The formula used is: [tex]S_n=\frac{1-r^n}{1-r}[/tex]
We have n =5, r=1/3
[tex]S_n=\frac{a(1-r^n)}{1-r}\\S_5=\frac{8(1-\frac{1}{3}^5)}{1-\frac{1}{3} }\\S_5=\frac{8(1-\frac{1}{243})}{1-\frac{1}{3} }\\S_5=\frac{8(\frac{243-1}{243})}{\frac{3-1}{3} }\\S_5=\frac{8(\frac{242}{243})}{\frac{2}{3} }\\S_5=\frac{8(0.995)}{0.66}\\S_5=12.06[/tex]
So, The sum of the geometric series is 12.06
