Answer:
The sum of the series is 9.99
Step-by-step explanation:
We are given the series [tex]\sum_{n=3}^{12}20(0.5)^{n-1}[/tex].
So, [tex]a_{n}=20(0.5)^{n-1}[/tex], where n=3 to 12
Then, we have,
[tex]1.\ a_{3}=20(0.5)^{3-1}\\a_{3}=20(0.5)^{2}\\a_{3}=5[/tex]
[tex]2.\ a_{4}=20(0.5)^{4-1}\\a_{4}=20(0.5)^{3}\\a_{4}=2.5[/tex]
Thus, the common ratio is [tex]r=\frac{2.5}{5}=0.5[/tex]
Since, the sum of first n terms of a series is [tex]S=\frac{a_{1}(1-r^{n})}{1-r}[/tex].
As n = 3 to 12, then the number of terms = 10, first term=a₃= 5 and r= 0.5
So, the sum of 10 terms is [tex]S=\frac{5(1-(0.5)^{10})}{1-0.5}[/tex]
i.e. [tex]S=\frac{5(1-0.00098)}{0.5}[/tex]
i.e. [tex]S=\frac{5\times 0.99902}{0.5}[/tex]
i.e. [tex]S=\frac{4.9951}{0.5}[/tex]
i.e. S = 9.99
Hence, the sum of the series is 9.99
