Given:
[tex]\sum_{a\mathop{=}1}^{11}64\cdot(0.2)^{a-1}[/tex]
To find: Sum of first 5 terms
Explanation:
Here, the first term is,
[tex]a_1=64[/tex]
The common ratio, r = 0.2
The number of terms n = 5.
Using the sum formula for geometric series,
[tex]S_n=\frac{a_1(1-r^n)}{1-r}[/tex]
Substituting the values we get,
[tex]\begin{gathered} S_5=\frac{64(1-0.2^5)}{1-0.2} \\ =\frac{64(1-0.2^{5})}{1-0.2} \\ =79.97 \end{gathered}[/tex]
Thus, the sum of the first 5 terms is 79.97.
Final answer:
[tex]S_5=79.97[/tex]
