1. **State the problem:** Find the sum of the first 60 terms of the geometric sequence \(\frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \ldots\). 2. **Identify the first term and common ratio:** - First term \(a = \frac{1}{3}\) - Common ratio \(r = \frac{1}{9} \div \frac{1}{3} = \frac{1}{3}\) 3. **Formula for the sum of the first \(n\) terms of a geometric sequence:** $$ S_n = a \frac{1-r^n}{1-r} $$ where \(a\) is the first term, \(r\) is the common ratio, and \(n\) is the number of terms. 4. **Apply the formula:** $$ S_{60} = \frac{1}{3} \times \frac{1 - \left(\frac{1}{3}\right)^{60}}{1 - \frac{1}{3}} $$ 5. **Simplify the denominator:** $$ 1 - \frac{1}{3} = \frac{2}{3} $$ 6. **Rewrite the sum:** $$ S_{60} = \frac{1}{3} \times \frac{1 - \left(\frac{1}{3}\right)^{60}}{\frac{2}{3}} = \frac{1}{3} \times \frac{3}{2} \left(1 - \left(\frac{1}{3}\right)^{60}\right) $$ 7. **Simplify:** $$ S_{60} = \frac{1}{2} \left(1 - \left(\frac{1}{3}\right)^{60}\right) $$ 8. **Interpretation:** Since \(\left(\frac{1}{3}\right)^{60}\) is an extremely small positive number, the sum is very close to \(\frac{1}{2}\). **Final answer:** $$ S_{60} = \frac{1}{2} \left(1 - \left(\frac{1}{3}\right)^{60}\right) $$