Problem 1226

1. The formula for the sum of an infinite geometric series is given by:
$$S = \frac{a}{1 - r}$$
where $a$ is the first term and $r$ is the common ratio (with $|r| < 1$). According to the problem, the sum of the series is 7:
$$\frac{a}{1 - r} = 7 \quad \text{(1)}$$

2. The sum of the terms involving odd powers of $r$ in an infinite geometric series is given by:
$$S_{\text{odd}} = \frac{ar}{1 - r^2}$$
According to the problem, this sum is 3:
$$\frac{ar}{1 - r^2} = 3 \quad \text{(2)}$$

3. We can express equation (2) in terms of equation (1). From equation (1), we have:
$$\frac{a}{1 - r} = 7 \implies a = 7(1 - r)$$
Substitute $a$ into equation (2):
$$\frac{7(1 - r)r}{1 - r^2} = 3$$

4. Simplify the equation:
$$\frac{7r(1 - r)}{(1 - r)(1 + r)} = 3$$
$$\frac{7r}{1 + r} = 3$$

5. Solve for $r$:
$$7r = 3(1 + r)$$
$$7r = 3 + 3r$$
$$4r = 3$$
$$r = \frac{3}{4}$$

6. Substitute $r = \frac{3}{4}$ back into equation (1) to find $a$:
$$\frac{a}{1 - \frac{3}{4}} = 7$$
$$\frac{a}{\frac{1}{4}} = 7$$
$$4a = 7$$
$$a = \frac{7}{4}$$

7. Finally, calculate $a + r$:
$$a + r = \frac{7}{4} + \frac{3}{4} = \frac{10}{4} = \frac{5}{2}$$

Conclusion:
$$a + r = \boxed{\frac{5}{2}}$$