The ratio of wavelengths of the last line of Balmer series and the last line of Lyman series is

The wavelength of last line of Balmer series 

$\frac{1}{{\lambda }_B}=Rc\left(\frac{1}{2^2}-\frac{1}{{\infty }^2}\right)=\frac{Rc}{4}$

The wavelength of last line of Lyman series 

$\frac{1}{{\lambda }_B}=Rc\left(\frac{1}{1^2}-\frac{1}{{\infty }^2}\right)=Rc$

$\therefore \frac{{\lambda }_B}{{\lambda }_L}=\frac{4}{1}=4$