What are the first term and common ratio given two facts about this geometric progression?

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The terms of our sequence can be written as \(a,ar,ar^2,...\) and the given facts tell us that \[\begin{align*} a&=ar+2 \\ ar+ar^2&=\frac{4}{3}. \end{align*}\] Rearranging and factorising, these equations become \[\begin{align*} a(1-r) &= 2 \\ ar(1+r) &=\frac{4}{3}. \end{align*}\] We can then divide our second equation by our first giving \[\frac{r(1+r)}{1-r}=\frac{2}{3}\] which we can rearrange into a quadratic. \[\begin{align*} 3r(1+r)&=2(1-r) \\ 3r^2+5r-2&=0 \end{align*}\]

Using the quadratic formula or by factorisation we find that the common ratio \[r=\frac{1}{3}\quad\text{or}\quad-2.\]

Now that we have the common difference we can calculate our first term \[a=\frac{2}{1-r}=3\quad\text{or}\quad\frac{2}{3}.\]

If all the terms are positive the common ratio cannot be negative, so we must have \(r=\frac{1}{3}\) and \(a=3\).

Substituting these values into the formula for the sum to infinity we have \[\begin{align*} S_\infty &= \frac{a}{1-r} \\ &=\frac{3}{2/3}=\frac{9}{2}. \end{align*}\]