Diketahui deret geometri u1,u2,u3+.... +u6 . Suku keenam u6 = 162 dan log u2 + log u3 + log u4 + log u5 = 4 log 2 + 6 log 3. Tentukan suku pertama serta rasio d

u6 = ar^5 = 162
r^5 = 162/a
log U2 + log U3 + log U4 + log U5 = 4.log 2 + 6.log 3
log (U2.U3.U4.U5) = 2(2.log 2 + 3.log 3)
log (ar.ar².ar³.ar⁴) = 2(log 2² + log 3³)
log (a⁴.r^10) = 2(log 4 + log 27)
log (a².r^5)² = 2(log (4×27))
2.log (a².r^5) = 2.log 108
log (a².r^5) = log 108
a².r^5 = 108
a²(162/a) = 108
a(162) = 108
a = 108/162
a = 2/3 (suku pertama)
r^5 = 162/a
r^5 = 162/(2/3)
r^5 = 162 × 3/2
r^5 = 243
r = 3 (rasio)