Find the sum of the first six terms of the geometric series in which a3 is -18 and a6 is 486.

An=a1*(r)^(n-1)

a3=-18=a1*(r)^(3-1)=a1*r^2
a6=486=a1*(r)^(6-1)=a1*r^5

a6/a3=(a1*r^5)/(a1*r^2)=r^3=486/-18=-27

r^3=-27
r=-3

math
-18=a1*r^2
-18=a1*(-3)^2
-18=a1*9
-2=a1

sum of first n terms of geo is

r=-3
a1=-2
n=6




364 is the sum of first 6 terms

Then why is the question on here if you know the answer

Idon’t exactly understand but i think dbc=rst