Tentukan jumlah 5 suku pertama pada deret geometri 3+3/2=3/4+……….

Diketahui :
a = U₁
b = U₂ – U₁
r  = U₂/U₁

Un = a + (n – 1)b.
Sn = n/2.(2a + (n – 1)b)

Ditanya :
S₅ = ???

Dijawab :
⇔ n = 5

⇔ a = U₁
       = 3

⇔ b = 3/2 – 3
⇔ b = 3/2 – 6/2
⇔ b = -3/2

⇔ r = U₂/U₁
⇔ r = (3/2)/3
⇔ r = 3/(2.3)
⇔ r = 3/6
⇔ r = 1/2

deret aritmatika
Sn = n/2.(2a + (n – 1)b)
S₅  = (5/2).(2.3 + (5-1).(-3/2)
     = (5/2).(6+ (4.-3/2))
     = (5/2).(6+(-12/2))
     = (5/2).(6-6)
     = (5/2).(0)
     = 0

deret geometri
S₅  = (3*(1-(1/2)^(5))/(1-(1/2))
S₅  = (3*(-(1/2)^(5))/(1/2)
S₅  = (3*(-(1/32))/(1/2)
S₅  = (3*2)(-(1/32)+1)/1
S₅  = (6)((32/32)-(1/32))
S₅  = (6)(31/32)
S₅  = (186/32)
S₅  = (93/16)

Jadi jumlah 5 suku pertama pada deret geometri tersebut adalah 93/16