1. **State the problem:** Simplify the expression $$\left(\frac{1}{5a} + \frac{1}{7a}\right) \times \frac{a^2}{4}$$. 2. **Find a common denominator for the sum inside the parentheses:** The denominators are $5a$ and $7a$. The least common denominator (LCD) is $35a$. 3. **Rewrite each fraction with the LCD:** $$\frac{1}{5a} = \frac{7}{35a}, \quad \frac{1}{7a} = \frac{5}{35a}$$ 4. **Add the fractions:** $$\frac{7}{35a} + \frac{5}{35a} = \frac{7+5}{35a} = \frac{12}{35a}$$ 5. **Multiply the sum by $\frac{a^2}{4}$:** $$\frac{12}{35a} \times \frac{a^2}{4} = \frac{12 \times a^2}{35a \times 4} = \frac{12a^2}{140a}$$ 6. **Simplify the fraction:** Cancel $a$ in numerator and denominator: $$\frac{12a^2}{140a} = \frac{12a}{140}$$ 7. **Reduce the fraction $\frac{12}{140}$ by dividing numerator and denominator by 4:** $$\frac{12a}{140} = \frac{3a}{35}$$ **Final answer:** $$\boxed{\frac{3a}{35}}$$