Geometric series... stuck at Q worth 9 marks helppp!!

The first, third and fifth term of a geometric series G are (x-1), 3x and (10x+8) respectively.
a) Find the possible values of x.
Given that all the terms of G are positive, find
b) The first term of G ( i know this )
c) The common ratio of G
The sum of the first n terms of G is S(n)
d) Find the Value of S(8)

If common ratio is r then.... (x-1), r(x-1), r^2( x-1) =3x, r^3( x-1) and r^4( x-1) = (10x+8)....are first 5 terms .

r^2 = 3x/(x-1), r^4 = (10x+8)/(x-1)

so, [3x/(x-1)]^2 = (10x+8)/(x-1)

9x^2 = (10x+8)(x-1) = 10x^2 -2x -8.......x^2 -2x -8 =0 ....( x-4) ( x+2) =0.

so, x =4, or x =-2

for all all positive values....first term is 4-1 =3,

common ratio r by........r.^2 = 3x/(x-1)=12/3 =4 ...... r =2

Sn = a(1-r^n)/[1-r]

here a =3, r=2

Sn = 3( 1-2^8)/(-1) =765

The 3rd term is r^2 * the 1st term.
The 5th term is r^2 * the 1st term.
You don't know what r is, but you know that the ratio a_3/a_1 = a_5/a_3 = r^2.

So you can set 3x/(x-1) = (10x + 8)/3x. Cross multiply and you'll have a quadratic equation. At most 2 solutions.

The phrase "given that all the terms of G are positive" tells me that there are two solutions for x and one of them is negative. You're supposed to choose the positive one for parts b)-d).

The first term of G is x - 1. How can you know this if you haven't found x yet?
Evaluate the 1st and 3rd terms, take the ratio a_3/a_1. That's r^2.
Use the formula for the sum of a geometric series, now that you know a_1 and r.