Tentukan Jumlah 6 suku pertama pada deret geometri a) 128 - 64 - 32 - 16 + ... b) 1 - 3/4 + 9/16 - 27/64 + ...

a.
a = 128
r = -1/2
[tex]\displaystyle S_6=\frac{a(1-r^6)}{1-r}=\frac{128(1-(-\frac{1}{2})^6)}{1-(-\frac{1}{2})} \\ S_6=\frac{128(1-\frac{1}{64})}{\frac{3}{2}}=\frac{128\times\frac{63}{64}}{\frac{3}{2}} \\ S_6 = \frac{126}{\frac{3}{2}}=126\times\frac{2}{3}=84[/tex]

b.
a = 1
r = -3/4
[tex]\displaystyle S_6=\frac{a(1-r^6)}{1-r}=\frac{1(1-(-\frac{3}{4})^6)}{1-(-\frac{3}{4})} \\ S_6=\frac{1-\frac{729}{4096}}{\frac{7}{4}}=\frac{\frac{3367}{4096}}{\frac{7}{4}} \\ S_6 = \frac{481}{1024}[/tex]