Which transition in the hydrogen spectrum would have the same wavelength as the Balmer type transition from $\mathrm{n=4}$ to $\mathrm{n}=2$ of $\mathrm{He}^{+}$ spectrum

$\mathrm{He}^{+}$ion : <br/><br/>$$ \begin{aligned} & \frac{1}{\lambda(\mathrm{H})}=\mathrm{R}(1)^2\left[\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right] \\\\ & \frac{1}{\lambda\left(\mathrm{He}^{+}\right)}=\mathrm{R}(2)^2\left[\frac{1}{2^2}-\frac{1}{4^2}\right] \\\\ & \text { Given } \lambda(\mathrm{H})=\lambda\left(\mathrm{He}^{+}\right) \\\\ & \mathrm{R}(1)^2\left[\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right]=\mathrm{R}(4)\left[\frac{1}{2^2}-\frac{1}{4^2}\right] \\\\ & \frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}=\frac{1}{1^2}-\frac{1}{2^2} \end{aligned} $$ <br/><br/>On comparing $\mathrm{n}_1=1 $ and $ \mathrm{n}_2=2$.