Tentukan jumlah deret geometri berikut a) 5 + 10 + 20 + ... + 320 b) 1 - 1 per 2 + 1 per 4 - ... + 1 per 64

a.
a = 5
b = 5
Un = 5n
Un = 320
5n = 320
n = 64
Maka,
S64 = 64/2 (5+320)
S64 = 32.325 = 10400

b.
1-1/2+1/4-...+1/64
a = 1
r = -1/2
[tex]U_n=ar^{n-1} \\ \frac{1}{64}=1.(-\frac{1}{2})^{n-1} \\ (-2)^{n-1}=64 \\ (-2)^{n-1}=(-2)^6 \\ n-1=6 \\ n=7 \\ \\ \\ \displaystyle S_7=\frac{a(1-r^7)}{1-r}=\frac{1(1-(-\frac{1}{2})^7)}{1-(-\frac{1}{2})} \\ S_7 = \frac{1+\frac{1}{128}}{\frac{3}{2}}=\frac{\frac{129}{128}}{\frac{3}{2}}=\frac{43}{64}[/tex]