Question

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

Answer 1

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

`(a_1(1-r^n))/(1-r)`
r=-3
a1=-2
n=6

`(-2(1-(-3)^6))/(1-(-3))`

`(-2(1-729))/(1+3)`

`(-2(-728))/(4)`

`(1456)/(4)`
364 is the sum of first 6 terms