dalam suatu deret goemetri, jumlah tiga suku pertamanya ialah 48 dan jumlah tiga berikutnya adalah -6. jumlah sampai tak hingga deret geometri itu adalah...

D3 (baris 1) = 48
D3 (baris 2) = -6

D3 (baris 1)
a + ar + ar^2 = 48
a(1 + r + r^2) = 48

D3 (baris 2)
ar^3 + ar^4 + ar^5 = -6
a(r^3 + r^4 + r^5) = -6

[a(r^3 + r^4 + r^5)]/[a(1 + r + r^2)] = -6/48
r^3 = -1/8
r^3 = (-1/2)^3
r = -1/2

a(1 + r + r^2) = 48
a(1 + (-1/2) + (-1/2)^2) = 48
a(1 - 1/2 + 1/4) = 48
a(4/4 - 2/4 + 1/4) = 48
a(3/4) = 48
a = 48/(3/4)
a = 48 × 4/3
a = 64


D~ = a/(1 - r)
D~ = 64/(1 - (-1/2))
D~ = 64/(1 + 1/2)
D~ = 64/(2/2 + 1/2)
D~ = 64/(3/2)
D~ = 64 × 2/3
D~ = 128/3
D~ = 42 2/3

Jadi jumlah tak hingga deret geometri itu adalah 42 2/3