In a geometric progression, the sum to infinity is 12 and the sum of the first two terms is 20/3. Find the first term and the common ratio.

The sum of the first $n$ terms is $S_n = \frac{a(1-r^n)}{1-r}$, where $a$ is the first term and $r$ is the common ratio.
The sum to infinity is $S_\infty = \frac{a}{1-r}$, valid for $|r|<1$.

Given that the sum to infinity is 12:

Given that the sum of its first terms is $\frac{20}{3}$. This refers to the sum of the first two terms, $S_2$.

From equation (1), express $a$ in terms of $r$:

Substitute equation (3) into equation (2):

Divide both sides by 12:

Simplify the fraction:

Rearrange to solve for $r^2$:

Take the square root of both sides:

The problem states that the geometric progression has positive terms. If $r = -\frac{2}{3}$, the terms would alternate in sign (e.g., $a, ar, ar^2, \dots$). For all terms to be positive, $r$ must be positive. Also, for $S_\infty$ to exist, $|r|<1$, which is satisfied by both $\frac{2}{3}$ and $-\frac{2}{3}$.
Therefore, we choose the positive value for $r$:

Now substitute $r = \frac{2}{3}$ back into equation (3) to find $a$:

Since $a=4$ and $r=\frac{2}{3}$ are both positive, all terms in the progression will be positive.

The first term is $\boxed{a=4}$ and the common ratio is $\boxed{r=\frac{2}{3}}$.

$a=4$ and $r=\frac{2}{3}$.

Using the formula $S_n = \frac{a(1-r^n)}{1-r}$:

The sum of the first three terms is $\boxed{\frac{76}{9}}$.