jmc

This is a geometric sequence with first term $1$ and common ratio $\frac{1}{3}$. Thus the sum of the first $n$ terms is: $$\frac{121}{81}=\frac{1\left[1-{\left(\frac{1}{3}\right)}^n\right]}{1-\frac{1}{3}}=\frac{3^n-1}{2\cdot 3^{n-1}}.$$Solving we have: $$\begin{array}{rl}\frac{3^n-1}{2\cdot 3^{n-1}}& =\frac{121}{81}\\ \frac{3^n-1}{2\cdot 3^n}& =\frac{121}{243}\\ 243(3^n-1)& =121\cdot 2\cdot 3^n\\ 243\cdot 3^n-243& =242\cdot 3^n\\ 3^n& =243\\ n& =\boxed{5}.\end{array}$$