\(\frac{1}{\lambda_1}=R(1)^2\left[\frac{1}{(2)^2}-\frac{1}{(3)^2}\right]=R\left(\frac{5}{36}\right)\)
\(\frac{1}{\lambda_2}=R(1)^2\left[\frac{1}{(3)^2}-\frac{1}{(4)^2}\right]=R\left(\frac{7}{144}\right)\)
\((ii) \div (i)\) gives
\(\frac{\lambda_1}{\lambda_2}=\frac{7 / 144}{5 / 36}=\frac{7}{20}=\frac{7}{4 \times 5}\)
\(\therefore n =5\)