I know this is an old question, but @Jdip answers it directly in this answer to my related question. I'll quote directly from his answer:
Digital Spectra Of Non-Uniformly Sampled Signals : Theories and
Applications is a somewhat obscure reference (from 1988! can't go
wrong!) that gives a general representation of the spectrum of a
special case of non-uniformly sampled digital signals: periodically
non-uniform sampled signal.
He continues with details from the paper:
The paper deals with a special case of non-uniform sampling,
Periodically nonuniform sampling, expanding on @Fats32’s answer regarding
sampling instants periodically repeated by blocks of $N$
It starts with a signal $g(t)$, band-limited to ($-1/2T, 1/2T$), with
Fourier transform $G^a(\omega)$:
The signal $g(t)$ is sampled in such a structured way that the sampling time instances [...] have an overall period $MT$.
It then defines another band-limited sequence, $\bar{g}(t)$:
The sampled data sequence is then treated as if it were obtained by sampling another function $\bar{g}(t)$, [...] at uniform rate $1/T$.
and states the aim of the paper (which answers your question, at
least for these types of non-uniformly sampled signals):
We are interested in finding the representation of the digital spectrum of $\bar{g}(t)$ in terms of the Fourier transform
$G^a(\omega)$ of $g(t)$.
Using the framework described above, it then goes through a really
nice derivation of $G(\omega)$ to arrive at two representations, from
which you can see the effect of the non-uniform sampling on the
spectrum:
$$G(\omega) = \frac{1}{MT}
\sum_{m=0}^{M-1}\left[\sum_{k=-\infty}^{\infty}G^a\left[\omega-k(\frac{2\pi}{MT})\right]e^{j[\omega-k(\frac{2\pi}{MT})]t_m}\right]e^{-jm\omega
T} \tag{2}$$
And re-writing (2) with $t_m = mT - r_mT$, $r_m$ being the ratio of
$mT - t_m$ to the average sampling period $T$ gives (4):
$$G(\omega) = \frac{1}{T}\sum_{k=-\infty}^{\infty}\left(
\frac{1}{M}\sum_{m=0}^{M-1}e^{-j\left[\omega-k(\frac{2\pi}{MT})\right]r_mT}e^{-jkm(\frac{2\pi}{M})}\right)G^a\left[\omega-k(\frac{2\pi}{MT})\right]
\tag{4}$$