Respuesta :
The series is a geometric series which is governed by the following formula:
[tex]\begin{gathered} T_n=ar^{n-1} \\ \text{where:} \\ T_n=nth\text{ term of the series} \\ a=\text{ first term of the series} \\ r=\text{common ratio of the series} \end{gathered}[/tex]Thus, we have that:
a = 3 ,
To find r, we have to solve for it as follows:
[tex]\begin{gathered} \text{since }T_n=ar^{n-1} \\ \text{Thus, the second term, }T_2\text{ is:} \\ T_2=ar^{2-1} \\ \Rightarrow T_2=ar^1 \\ \sin ce\text{ }T_2=\frac{15}{4},\text{ and a =3. we have:} \\ \Rightarrow T_2=ar \\ \frac{15}{4}=3\times r \\ 3\times r=\frac{15}{4} \\ \Rightarrow r=\frac{\frac{15}{4}}{3}=\frac{15}{4}\times\frac{1}{3}=\frac{5}{4} \\ \Rightarrow r=\frac{5}{4} \end{gathered}[/tex]Thus, r = 5/4
- Now the sum of the first n terms of a geometric series is given by the formula:
[tex]S_n=\frac{a(r^n-1)}{r-1}[/tex]Thus, the sum of the first 9 terms is:
[tex]S_9=\frac{3\times(\frac{5}{4}-1)}{\frac{5}{4}-1}[/tex]simplifying the above gives:
[tex]\begin{gathered} S_9=\frac{3\times((\frac{5}{4})^9-1)}{\frac{5}{4}-1} \\ \Rightarrow S_9=\frac{3\times(\frac{1953125}{262144}-1)}{\frac{5}{4}-\frac{4}{4}}=\frac{3\times(\frac{1953125-262144}{262144})}{\frac{5-4}{4}}=\frac{3\times\frac{1690981}{262144}}{\frac{1}{4}} \\ \Rightarrow S_9=3\times4\times\frac{1690981}{262144}=\frac{20291772}{262144}=77.4 \\ \Rightarrow S_9=77\text{ (to the nearest integer)} \end{gathered}[/tex]Therefore, the sum of the first 9 terms of the series is 77 (to the nearest integer)
