In hydrogen spectrum the ratio of wavelengths of the last line of Lyman series and the last line of Balmer series is

The wavelength of a spectral line in the hydrogen spectrum is given by Rydberg's formula:

$\dfrac{1}{\lambda} = R \left( \dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} \right)$

For the last line of the Lyman series, the transition is from $n_2 = \infty$ to $n_1 = 1$.

$\dfrac{1}{\lambda_L} = R \left( \dfrac{1}{1^2} - \dfrac{1}{\infty^2} \right) = R$

$\lambda_L = \dfrac{1}{R}$

For the last line of the Balmer series, the transition is from $n_2 = \infty$ to $n_1 = 2$.

$\dfrac{1}{\lambda_B} = R \left( \dfrac{1}{2^2} - \dfrac{1}{\infty^2} \right) = \dfrac{R}{4}$

$\lambda_B = \dfrac{4}{R}$

The ratio of the wavelengths is:

$\dfrac{\lambda_L}{\lambda_B} = \dfrac{\dfrac{1}{R}}{\dfrac{4}{R}} = \dfrac{1}{4} = 0.25$

Answer: $0.25$