Given:
a1 = 4
a3 = 1
r = ½
Let's find the S3 of the sum of the geometric series.
Apply the sum of geometric series formula below:
[tex]S_n=\frac{a_1(1-r^n)}{1-r}[/tex]Let's solve for S3.
Substitute the values into the equation.
Where: n = 3
Thus, we have:
[tex]S_3=\frac{4(1-(\frac{1}{2})^3)^{}^{}}{1-\frac{1}{2}}[/tex]Solving further:
[tex]\begin{gathered} S_3=\frac{4(1-(\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}))}{\frac{1}{2}} \\ \\ S_3=\frac{4(1-\frac{1}{8})}{\frac{1}{2}} \\ \\ S_3=\frac{4(\frac{7}{8})}{\frac{1}{2}} \\ \\ S_3=\frac{4\ast\frac{7}{8}}{\frac{1}{2}} \\ \\ S_3=\frac{\frac{7}{2}}{\frac{1}{2}} \\ \\ S_3=\frac{7}{2}\ast\frac{2}{1} \\ \\ S_3=7 \end{gathered}[/tex]Therefore, the S3 of the sum of the given geometric series is 7
ANSWER:
7
