You use the next formula to get a explicit equation for a geometric sequence:
![a_n=a_1\cdot r^(n-1)](https://img.qammunity.org/2023/formulas/mathematics/college/usnb6cvy5q0c41ojuruucgvjnfnf10g7si.png)
In this case you use as the first data 2000 (term when n=1).
r is the common ratio between each term.
To find r you divide each term into the previous term as follow:
![\begin{gathered} (2000)/(1600)=(5)/(4) \\ \\ (2500)/(2000)=(5)/(4) \\ \\ (3125)/(2500)=(5)/(4) \\ \\ (3906.25)/(3125)=(5)/(4) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/bwl229a44dar545v5t4tr495uo861uq415.png)
Then, you get the next explicit equation: writen in two different forms
![\begin{gathered} t_n=2000\cdot((5)/(4))^(n-1) \\ \\ t_n=(2000\cdot5^(n-1))/(4^(n-1)) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/890us96ubk13vuwko4jjfnom19qj427x73.png)
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For the recursive formula you have the next:
Where tn-1 is the previus term
For the given sequence:
![\begin{gathered} t_1=200 \\ \\ t_n=(5)/(4)(t_(n-1)) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/zuj919mv9hz9jq8xnowlhxst37sxxu9qtv.png)