diketahui suku suku geometri positif u2= 14 dan u5 = 112.jumlah 6 suku pertama deret tersebut adalah

U2 = ar^(2-1)
ar = 14

U5 = ar^(5-1)
ar^4 = 112

ar×r^3 = U5
14 × r^3 = 112
r^3 = 112/14
r^3 = 8
r = 2

ar = U2
a×2 = 14
a = 14/2
a = 7

Sn = a(r^n - 1)/(r - 1)
S6 = 7(2^6 - 1)/(2 - 1)
S6 = 7(64 - 1)/1
S6 = 7(63)
S6 = 441