The shortest wavelength of H-atom in the Lyman series is  ${\lambda }_1$. The longest wavelength in the balmer series of  $H{e}^{+}$  is

 $\overline{\vartheta }=\frac{1}{\lambda }=R_H[\frac{1}{{n_1}^2}-\frac{1}{{n_2}^2}]Z^2$
Shortest wavelength means 
$n_2=\alpha n_1=1$  in Ly man series
 $\lambda ={\lambda }_1$
$\frac{1}{{\lambda }_1}=R_H[\frac{1}{1^2}-\frac{1}{{\alpha }^2}]=R$ 
 ${\lambda }_1=\frac{1}{R}$
For longest wavelength in Balmer series  $n_1=2, n_2=3, Z=2$
$\frac{1}{{\lambda }_2}=R_H\times Z^2[\frac{1}{{n_1}^2}-\frac{1}{{n_2}^2}]$

$$\begin{gathered} \frac{1}{{\lambda }_2}=R_H\times \frac{5}{9} \\ {\lambda }_2={\lambda }_1\times \frac{9}{5} \end{gathered}$$