Suatu barisan geometri diketahui a = 5, r = 3, dan Sn = 200. carilah n.

[tex]sn = \frac{a( {r}^{n} - 1) }{r - 1} \\ 200 = \frac{5( {3}^{n} - 1) }{3 - 1} \\200(2) = 5( {3}^{n} - 1) \\ \frac{400}{5} = {3}^{n} - 1 \\ 80 + 1 = {3}^{n} \\ 81 = {3}^{n} \\ {3}^{4} = {3}^{n} \\ n = 4[/tex]
semoga mebantu