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