Suatu barisan geometri mempunyai U7 = 16 dan U10 = 1024, maka U12 adalah....

Gunakan metode eliminasi.

[tex]u7 = {ar}^{6} = 16[/tex]

[tex]u10 = {ar}^{9} = 1024[/tex]

[tex] \frac{ {ar}^{9} }{ {ar}^{6} } = \frac{1024}{16} [/tex]

[tex] {r}^{3} = 64[/tex]

[tex]r = \sqrt[3]{64} [/tex]

[tex]r = 4[/tex]

[tex] {ar}^{6} = 16[/tex]

[tex]a(4) {}^{6} = 16[/tex]

[tex]a(4096) = 16[/tex]

[tex]a = \frac{16}{4096} [/tex]

[tex]a = \frac{1}{256} [/tex]

sudah diketahui r = 4 dan a = 1/256 , maka :

[tex]un = {ar}^{n - 1} \\ u12 = \frac{1}{256} \times 4 {}^{12 - 1} \\ u12 = \frac{1}{256} \times {4}^{11} \\ u12 = \frac{1}{ {2}^{8} } \times {2}^{22} \\ u12 = {2}^{14} [/tex]
[tex]u12 = 16384[/tex]