= Solution
Let the source's proper emission period be $T_0$, and let it approach a stationary observer directly with speed $v=\beta c>0$. In the observer's <inertial frame>, consecutive emissions are separated by $\gamma T_0$ through <time dilation>. The second crest starts $v\gamma T_0$ nearer the observer, reducing the arrival interval to
$$
T_{\rm obs}=\gamma T_0(1-\beta).
$$
Since light travels at $c$ in each <inertial frame>, the longitudinal <relativistic Doppler effect> is
$$
\boxed{\frac{\nu_{\rm obs}}{\nu_0}=\sqrt{\frac{1+\beta}{1-\beta}},\qquad
\frac{\lambda_{\rm obs}}{\lambda_0}=\sqrt{\frac{1-\beta}{1+\beta}}.}
$$
Recession reverses the sign of $\beta$. More generally, if $\Theta$ is the angle between the source velocity and the photon propagation direction in the observer's frame, a <Lorentz transformation> of the photon <four-momentum> gives $\nu_0=\gamma\nu_{\rm obs}(1-\beta\cos\Theta)$.
The <common Doppler scaling of spectral-line separations> follows because every line receives the same wavelength factor $D$. The supplied wavelengths have rest separation $50\,\mathrm{nm}$; the measured separation gives $D=51/50=1.02$. Defining $\beta_{\rm rec}>0$ for recession, inversion gives
$$
\boxed{\beta_{\rm rec}=\frac{D^2-1}{D^2+1}=\frac{101}{5101}=0.0198000,\qquad
v_{\rm rec}=5.93590\times10^6\,\mathrm{m\,s^{-1}}.}
$$
\b[Source contradiction:] a strictly approaching source has $D<1$ and must reduce the wavelength separation below $50\,\mathrm{nm}$, so no positive approach speed satisfies the stated $51\,\mathrm{nm}$. For example, approach at $0.02c$ gives a separation of $49.0098\,\mathrm{nm}$. The speed inferred from the numerical data is therefore about $0.0198c$ \b[receding]; retaining the word “approaching” would require a negative signed approach velocity. The two observed wavelengths consistent with the given separation are $357$ and $408\,\mathrm{nm}$.
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