Solution :
The shortest wavelength transition corresponds to $n_2 \infty$ to $n_1=2$ transition .
$\bar{v}=109678 \bigg[ \large\frac{1}{n_1^2}-\frac{1}{n_2^2}\bigg]$$cm^{-1}$
$\qquad= 109678 \bigg[\large\frac{1}{(2)^2}-\frac{1}{\infty}\bigg]$
$\qquad= \large\frac{109678}{4}$
$\qquad= 27419.5\;cm^{-1}$