In hydrogen spectrum, the shortest wavelength in the Balmer series...

Shortest wavelength means maximum energy in transition.

Therefore for Balmer series transition would be $n = \infty \text{to}n = 2$ and for Brackett series transition would be from $n = \infty \text{to}n = 4$.

Now from Rydberg's formula,

$\dfrac{\dfrac{1}{{\lambda}_{Bl}}}{\dfrac{1}{{\lambda}_{Br}}} = \dfrac{{R}_{H}\left(\dfrac{1}{{2}^{2}} - \dfrac{1}{{\infty}^{2}}\right)}{{R}_{H}\left(\dfrac{1}{{4}^{2}} - \dfrac{1}{{\infty}^{2}}\right)} ⇒ {\lambda}_{Br} = 4{\lambda}_{Bl} = 4\lambda$