If the shortest wavelength in Lyman series of hydrogen atom is A, then the longest wavelength in Paschen series of He⁺ is

For Lyman series hydrogen atom, the shortest wavelength corresponds to:
n ₁ =1 and n ₂ =∞

$$\begin{gathered} \frac{1}{\lambda }=R{Z}^2\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right] \\ \frac{1}{A}=R{Z}^2\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right] \\ \frac{1}{A}=R------------\left(i\right) \end{gathered}$$

For Paschen series of He+  ion, the longest wavelength corresponds to :

n ₁ =3 and n ₂ =4

$$\begin{gathered} \frac{1}{\lambda }=R{Z}^2\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right] \\ \frac{1}{\lambda }=\frac{1}{\lambda }{2}^2\left[\frac{1}{3_1^2}-\frac{1}{4_2^2}\right] \\ \lambda =\frac{36A}{7} \end{gathered}$$