#### Answer

The longest three wavelengths in the Lyman series are $121.5~nm$, $102.6~nm$, and $97.23~nm$

#### Work Step by Step

The longest three wavelengths in the Lyman series would be the three transitions with the least amount of energy, that is, the transitions from level 2 to the ground state, from level 3 to the ground state, and from level 4 to the ground state.
We can find the wavelength of the photon emitted by a transition from level 2 to the ground state:
$\frac{1}{\lambda} = R~(\frac{1}{n_f^2}-\frac{1}{n_i^2})$
$\frac{1}{\lambda} = R~(\frac{1}{1^2}-\frac{1}{2^2})$
$\frac{1}{\lambda} = R~(\frac{3}{4})$
$\lambda = \frac{4}{3~R}$
$\lambda = \frac{4}{(3)~(1.097\times 10^7~m^{-1})}$
$\lambda = 1.215\times 10^{-7}~m$
$\lambda = 121.5~nm$
We can find the wavelength of the photon emitted by a transition from level 3 to the ground state:
$\frac{1}{\lambda} = R~(\frac{1}{n_f^2}-\frac{1}{n_i^2})$
$\frac{1}{\lambda} = R~(\frac{1}{1^2}-\frac{1}{3^2})$
$\frac{1}{\lambda} = R~(\frac{8}{9})$
$\lambda = \frac{9}{8~R}$
$\lambda = \frac{9}{(8)~(1.097\times 10^7~m^{-1})}$
$\lambda = 1.026\times 10^{-7}~m$
$\lambda = 102.6~nm$
We can find the wavelength of the photon emitted by a transition from level 4 to the ground state:
$\frac{1}{\lambda} = R~(\frac{1}{n_f^2}-\frac{1}{n_i^2})$
$\frac{1}{\lambda} = R~(\frac{1}{1^2}-\frac{1}{4^2})$
$\frac{1}{\lambda} = R~(\frac{15}{16})$
$\lambda = \frac{16}{15~R}$
$\lambda = \frac{16}{(15)~(1.097\times 10^7~m^{-1})}$
$\lambda = 0.9723\times 10^{-7}~m$
$\lambda = 97.23~nm$
The longest three wavelengths in the Lyman series are $121.5~nm$, $102.6~nm$, and $97.23~nm$.