Wavelength $\lambda$ of spectral lines in $\mathrm{H}$-atom is given as
$$
\frac{1}{\lambda}=R\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right) \quad \text{...(i)}
$$
For shortest wavelength of Paschen series,
$$
n_{1}=3 \text { and } n_{2}=\infty
$$
$\therefore$ From Eq. (i), we get
$$
\begin{aligned}
& \frac{l}{\lambda_{P S}}=R\left(\frac{1}{3^{2}}-\frac{1}{\infty}\right)=\frac{R}{9} \\
\Rightarrow \quad & \lambda_{P S}=\frac{9}{R} \quad \text{...(ii)}
\end{aligned}
$$
For shortest wavelength in Balmer series,
$$
n_{1}=2, n_{2}=\infty
$$
$\therefore$ From Eq. (i), we get
$\frac{1}{\lambda_{B S}}=R\left(\frac{1}{2^{2}}-\frac{1}{\infty}\right)=\frac{R}{4}$
$\Rightarrow \quad \lambda_{B S}=\frac{4}{R} \quad \text{...(iii)}$
For shortest wavelength of Lyman series,
$$
n_{1}=1 \text { and } n_{2}=\infty
$$
$\therefore$ From Eq. (i), we get
$$
\begin{aligned}
\frac{1}{\lambda_{L S}} &=R\left(\frac{1}{1^{2}}-\frac{1}{\infty}\right)=R \\
\Rightarrow \quad \lambda_{L S} &=\frac{1}{R}
\end{aligned}
$$
Hence, from Eq. (i), (ii) and (iii), we have
$$
\begin{aligned}
\lambda_{P S}: \lambda_{B S}: \lambda_{L S}=& \frac{9}{R}: \frac{4}{R}: \frac{1}{R} \\
&=9: 4: 1
\end{aligned}
$$