Obtain an expression for the nth term of the Geometric Progression. The 2nd, 6th and 8th terms of an Arithmetic progression are three successive terms of a Geometric progression.
Find the common ratio of the Geometric progression and obtain an expression for the $n$th term of the geometric progression.
Working so far:
Let $x, xr$ and $xr^2$ be the three terms of the G.P..
Let $a$ be the first term for the A.P..
$\begin{align} a+d&=x\tag {1}\\a+5d&=xr\tag{2}\\a+7d&=xr^2 \tag{3}\end{align}$
$\begin{align}(2)-(1): 4d&=x(r-1)\tag{4}\\(3)-(2):2d&=xr(r-1)\tag{5}\\\frac{(5)}{(4)}: r&=\frac{1}{2}\end{align}$
So, I found the common ratio but I don’t know how to obtain an expression for the $n$th term of the G.P. because there are four unknowns (one already solved) but I only have 3 equations.
The answer from the book is $\frac {16}{9}a(\frac{1}{2})^n$ but I don’t know how to get the $\frac{16}{9}a$ . I’m assuming the $a$ given by the book is the first term of the A.P., that’s why I used $a$ for the first A.P. term and $x$ as the first term of G.P..

Working backwards from the answer,
$\begin{align}\frac {16}{9}a(\frac{1}{2})^n&=xr^{n-1}\\ \frac {16}{9}a(\frac{1}{2})^n&=\frac{x(\frac{1}{2})^n}{\frac{1}{2}}\\\frac{16}{9}a&=2x\\\frac{8}{9}a&=x\end{align}$
So, that would mean $x=\frac{8}{9}a$, where $x$ is the first term of the G.P..
However, taking $\frac{(2)}{(1)}$:
$\frac{a+5d}{a+d}=\frac{1}{2}$
$2a+10d=a+d$
$d=\frac{-1}{9}a$
Substituting $d=\frac{-1}{9}a$ into $(1)$:
$x=\frac{8}{9}a$
