How to solve this geometric series question?

Question

A geometric series has first term 4 and common ratio r, where 0< r<1. The first, second, and fourth terms of this geometric series form three successive terms of an arithmetic series. Find the sum to infinity of the geometric series.

$S = \frac{t_1}{1-r}$ is the sum to infinity where $t_1$ is the first term in the geometric series.

The second term of the arithmetic series is $t_1*r$, the third term is $t_1*r^3$. From there we can deduce that $(r-1)(r^2+r-1)=0.$ So r must be either $\frac{-1 - \sqrt{5}}{2}$ or $\frac{-1 + \sqrt{5}}{2}$. From there how can I find $t_1$ and the sum to infinity?

Thank you very much!

Answer 1

The question says the first term is $4$, so $t_1 = 4$. Thus, since $0 \lt r \lt 1$, you have $r = \frac{-1 + \sqrt{5}}{2}$ giving

$$S = \frac{t_1}{1-r} = \frac{4}{1 - \frac{-1 + \sqrt{5}}{2}} = \frac{8}{2 - (-1 + \sqrt{5})} = \frac{8}{3 - \sqrt{5}} = 2\left(3 + \sqrt{5}\right) \tag{1}\label{eq1}$$

Thanks to J. W. Tanner for suggesting providing the normalized value at the end.

Answer 2

Let 1st term be a=4 and common ratio = r 2nd term a2=4*r 4th term is a4= ar^3=4r^3 They are in AP 2*(4r)=4 + 4r^3 r^3 -2 r +1 =0 and roots are r=1 , (sqrt.(5)-1)/2 , ( -1- sqrt(5))/2

Since 0