$$x_a(t) = 10 + 4\cos(30\pi t +\frac\pi 3) - \sin(250\pi t + \frac 14) -3\cos(500t + \frac \pi4) $$
$\textbf{Question:}$
This signal is first sampled at a rate of $f_s = 100$ samples per second to get a discrete array $x[n]$. Find $x[n]$. That $x[n]$ is transmitted to the receiver. The receiver only knows that the received discrete signal is obtained from a continuous-time signal by sampling at a rate of 100 samples/s. In an attempt to recover the original continuous signal from the received discrete signal, the receiver converts the received discrete signal first to an impulse train using $T_s = 1/100$ seconds and then uses an ideal low-pass filter to generate the $x_r(t)$; the passband of the filter is $|\omega| < 100\pi$ rd/s, and its gain is 1/100. Find $x_r(t)$; make sure that it is in the form of a superposition of real valued sinusoids.
$\textbf{My Attempt:}$
I found $x[n]$ as: $$x[n]=10 + 4\cos(\frac{3\pi}{10}n + \pi/3) - \sin(\frac{5\pi}{2}n + 1/4) -\cos(5n + \pi/4)$$
I know that $\mathcal{X_p}(j\omega)$, which is obtained by taking the Fourier Transform of the original signal $x_a(t)$ multiplied with the impulse train, is equal to $\mathcal{X}(j\Omega)|_{\Omega = \omega T}$ which is the Discrete Time Fourier transform of $x[n]$.
So now in order to get $x_r(t)$ should I just evaluate $$\mathcal{X}(j\Omega)|_{\Omega = \omega T} \cdot \mathcal{H}_{LP}(j\omega) = X_r(j\omega)$$ and then take the inverse fourier transform of $X_r(j\omega)$ to get $x_r(t)$ or is there some other simpler way?