A geometric series has a second term of 6. The sum of its fist 3 terms is -14. Find the fourth term.
I started with the Sn formula: Sn=(x(r^n-1))/(r-1). Since the first term (x) = 6/r,
(1) -14= 6(r^3-1)/(r(r-1))
(2) -14= (6r^3-6)/(r^2-r)
(3) -14=(6r^2-6)/-r
(4) 6r^2-14r-6 .... the quadratic equation gave me r= 0.37 or -2.7
(5) The fourth number should be equal to 6r, meaning either 2.22 or -16.23
I know my math is wrong, and I think my mistake is in step (3). I would appreciate if someone could correct me.
I started with the Sn formula: Sn=(x(r^n-1))/(r-1). Since the first term (x) = 6/r,
(1) -14= 6(r^3-1)/(r(r-1))
(2) -14= (6r^3-6)/(r^2-r)
(3) -14=(6r^2-6)/-r
(4) 6r^2-14r-6 .... the quadratic equation gave me r= 0.37 or -2.7
(5) The fourth number should be equal to 6r, meaning either 2.22 or -16.23
I know my math is wrong, and I think my mistake is in step (3). I would appreciate if someone could correct me.