The sum of an infinite geometric series is 16/3 and the sum of the first two terms is 5. Find the common ratio.

Let the first term of the geometric series be $a$ and the common ratio be $r$.
The sum of an infinite geometric series is given by the formula:

$$S_{\infty} = \frac{a}{1-r} \quad \text{for } |r| < 1$$

The sum of the first two terms is $a + ar$.

Given that the sum of the infinite geometric series is $\frac{16}{3}$:

$$\frac{a}{1-r} = \frac{16}{3} \quad (1)$$

Given that the sum of the first two terms is 5:

$$a + ar = 5$$ $$a(1+r) = 5 \quad (2)$$

From equation (2), express $a$ in terms of $r$:

$$a = \frac{5}{1+r}$$

Substitute this expression for $a$ into equation (1):

$$\frac{\frac{5}{1+r}}{1-r} = \frac{16}{3}$$ $$\frac{5}{(1+r)(1-r)} = \frac{16}{3}$$

Using the difference of squares formula, $(1+r)(1-r) = 1-r^2$:

$$\frac{5}{1-r^2} = \frac{16}{3}$$

Cross-multiply:

$$3 \times 5 = 16 \times (1-r^2)$$ $$15 = 16 - 16r^2$$

Rearrange the equation to solve for $r^2$:

$$16r^2 = 16 - 15$$ $$16r^2 = 1$$ $$r^2 = \frac{1}{16}$$

Take the square root of both sides:

$$r = \pm \sqrt{\frac{1}{16}}$$ $$r = \pm \frac{1}{4}$$

Both values, $r = \frac{1}{4}$ and $r = -\frac{1}{4}$, satisfy the condition $|r| < 1$ for an infinite geometric series.

The common ratio of the series can be $\frac{1}{4}$ or $-\frac{1}{4}$.

The final answer is $\pm \frac{1}{4}$.