suatu deret geometri mempunyai suku kedua 6 dan suku kelima 48. jumlah 8 suku pertama deret tersebut adalah?

[tex] \frac{a {r}^{4} }{ar} = \frac{48}{6} \\ {r}^{3} = 8 \\ r = 2 \\ ar = 6 \\ 2a = 6 \\ a = 3 \\ s _{n} = \frac{a( {r}^{n} - 1)}{r - 1} \\ s _{8} = \frac{3( {2}^{8} - 1)}{2 - 1} \\ s _{8} = 765 [/tex]

...,6,...,...,48
48÷6=8
[tex] \sqrt[3]{8 } [/tex]
=2

6÷2,6,6×2,6×2×2,48=
3,6,12,24,48,48×2,48×2×2,48×2×2×2
3,6,12,24,48,96,192,384,768
3+6+12+24+48+96+192+384+768=
1533