suku ke 5 barisan geometri adalah 12, dan suku ke 2 adalah 96. Tentukan jumlah 8 suku pertama deret geometri?

Diketahui :
Barisan Geometri

[tex]U_5 = 12 = ar^4 \\ U_2 = 96 = ar \\ \frac{U_5}{U_2} = \frac{ar^4}{ar} = \frac{12}{96} \\ \frac{U_5}{U_2} = r^3 = \frac{1}{8} \\ r = \frac{1}{2} [/tex]

Substitusi ke U2
[tex] U_2 = ar = 96 \\ a(\frac{1}{2}) = 96 \\ a = 192 [/tex]

[tex] S_n = \frac{a(1-r^n)}{1-r} \\ S_8 = \frac{192(1-(\frac{1}{2})^8))}{1-\frac{1}{2}} \\ S_8 = \frac{192(1-\frac{1}{256})}{\frac{1}{2}} \\ S_8 = \frac{192(\frac{255}{256})}{\frac{1}{2}} \\ S_8 = \frac{765}{2} [/tex]