Explanation

(d) Shortest wavelength comes from $n_{1} = \infty$ to $n_{2} = 1$ and longest wavelength comes from $n_{1} = 6$ to $n_{2} = 5$ in the given case. Hence $\frac{1}{\lambda_{\min}} = R\left( \frac{1}{1^{2}} - \frac{1}{\infty^{2}} \right) = R$
$\frac{1}{\lambda_{\max}} = R\left( \frac{1}{5^{2}} - \frac{1}{6^{2}} \right) = R\left( \frac{36 - 25}{25 \times 36} \right) = \frac{11}{900}R$
$\therefore\frac{\lambda_{\max}}{\lambda_{\min}} = \frac{900}{11}$