suatu deret geometri memiliki suku pertama 12 dan rasio 1/2. jumlah lima suku pertama deret tersebut adalah? Beserta caranya gimana

Sn = a(1-r^n)/1-r
=12 (1-1/2^5)/1-1/2
=12(1-1/32)per1/2
=12(31/32)per 1/2
=372/32 x 2/1
=23,25

trims

a=12
r=1/2
Sn= [tex] \frac{(1-r ^{n} )}{1-r}
S5= \frac{12(1- (\frac{1}{2 })^{2}) }{ \frac{1}{2} } = \frac{12- \frac{12}{4} }{1- \frac{1}{2} } = \frac{ \frac{48}{4}- \frac{12}{4} } { \frac{2}{2} -\frac{1}{2} } = \frac{ \frac{36}{4} }{ \frac{1}{2} } = \frac{36}{4} * \frac{2}{1} = 18[/tex]