The formula used to calculate the nth term of a geometric sequence is given to be:
![a_n=a_1\cdot r^(n-1)](https://img.qammunity.org/2023/formulas/mathematics/college/usnb6cvy5q0c41ojuruucgvjnfnf10g7si.png)
From the question, we are given the following parameters:
![\begin{gathered} a_1=34 \\ r=-(1)/(2) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/yc85j773diui3kb42fm56ech9fenpk4hvl.png)
Therefore, we can calculate the first 5 terms as follows:
First Term: 34
Second Term: -17
![\begin{gathered} n=2 \\ \therefore \\ a_2=34(-(1)/(2))^(2-1)=34*(-(1)/(2)) \\ a_2=-17 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/m0f5wir7if2sluhzl66ktqxe0iuv0v105a.png)
Third Term: 8.5
![\begin{gathered} n=3 \\ \therefore \\ a_3=34(-(1)/(2))^(3-1)=34*(1)/(4) \\ a_3=8.5 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/gm2yjovbd53u2ltq5uisnxoqzsd771ep0w.png)
Fourth Term: -4.25
![\begin{gathered} n=4 \\ \therefore \\ a_4=34(-(1)/(2))^(4-1)=34*(-(1)/(2))^3=34*(-(1)/(8)) \\ a_4=-4.25 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/2ezd1l06y7dllgjic6fv7ltqwqhiyroutd.png)
Fifth Term:
![\begin{gathered} n=5 \\ \therefore \\ a_5=34(-(1)/(2))^(5-1)=34*(-(1)/(2))^4=34*(1)/(16) \\ a_5=2.125 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/b2bvhlj1ld2szlpy4ihprdd2ib3lo5u2ly.png)
The first five terms are 34, -17, 8.5, -4.25, and 2.125.