1. Use the Rydberg formula for the Paschen series

   For hydrogen, $\frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$ In the Paschen series, the final level is $n_1=3$

2. First line of Paschen series

   The first line corresponds to transition: $n_2=4 \to n_1=3$

   So,

   =R\left(\frac{1}{9}-\frac{1}{16}\right)$$ Given: $$\lambda_1=720\text{ nm}$$

3. Second line of Paschen series

   The second line corresponds to transition: $n_2=5 \to n_1=3$

   So, $\frac{1}{\lambda_2}=R\left(\frac{1}{9}-\frac{1}{25}\right)$

4. Take ratio to eliminate $R$

   \frac{\left(\frac{1}{9}-\frac{1}{25}\right)}{\left(\frac{1}{9}-\frac{1}{16}\right)}$$ Compute both terms: $$\frac{1}{9}-\frac{1}{25}=\frac{25-9}{225}=\frac{16}{225}$$ $$\frac{1}{9}-\frac{1}{16}=\frac{16-9}{144}=\frac{7}{144}$$ Hence, $$\frac{1/\lambda_2}{1/\lambda_1}=\frac{16/225}{7/144} =\frac{16}{225}\cdot\frac{144}{7}$$ $$=\frac{16\cdot 144}{225\cdot 7}$$ Therefore, $$\lambda_2=\lambda_1\cdot\frac{\left(\frac{1}{9}-\frac{1}{16}\right)}{\left(\frac{1}{9}-\frac{1}{25}\right)}$$ $$\lambda_2=720\cdot\frac{7/144}{16/225}$$ $$=720\cdot\frac{7}{144}\cdot\frac{225}{16}$$

5. Simplify

   $\frac{720}{144}=5$

   So, $\lambda_2=5\cdot 7\cdot \frac{225}{16}$

   $=35\cdot \frac{225}{16}$

   $=\frac{7875}{16}=492.1875\text{ nm}$

6. Nearest integer

   $\lambda_2\approx 492\text{ nm}$

Final Answer: $492$ nm