The ratio of minimum wavelengths of Lyman and Balmer series will be

The series end of Lyman series corresponds to transition from $n_i=\infty$ to

$n_f=1$, corresponding to the wavelength

$\frac{1}{{\left({\lambda }_{\mathrm{m}\mathrm{i}\mathrm{n}}\right)}_{\mathrm{L}}}$ $=$ $R\left[\frac{1}{1}-\frac{1}{\infty }\right]$ $= R$

$\Rightarrow \mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }{\left({\mathrm{\lambda }}_{\mathrm{m}\mathrm{i}\mathrm{n}}\right)}_{\mathrm{L}}$ $=\frac{1}{R}$ $=$ 912 $Å$ ...(i)

For last line of Balmer series

$\frac{1}{{\left({\mathrm{\lambda }}_{\mathrm{m}\mathrm{i}\mathrm{n}}\right)}_{\mathrm{B}}}=R\left[\frac{1}{{\left(2\right)}^2}-\frac{1}{{\left(\infty \right)}^2}\right]=\frac{R}{4}$

$\Rightarrow \mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }{\left({\mathrm{\lambda }}_{\mathrm{m}\mathrm{i}\mathrm{n}}\right)}_{\mathrm{B}}=\frac{4}{R}=364\mathrm{8 }Å$ ...(ii)

Dividing Eq.(i) by Eq. (ii) .we get

$\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\frac{{\left({\mathrm{\lambda }}_{\mathrm{m}\mathrm{i}\mathrm{n}}\right)}_{\mathrm{L}}}{{\left({\mathrm{\lambda }}_{\mathrm{m}\mathrm{i}\mathrm{n}}\right)}_{\mathrm{B}}}=\text{0.25}$