Consider the DTFT of an N-sequence $x_n$:
$$ \begin{align*} X(e^{j\omega}) &= \sum_{n=0}^{N-1} x_n e^{-j\omega n} & \text{(DTFT)} \\ &= \sum_n \left ( \frac{1}{N} \sum_{k=0}^{N-1} X_k W_N^{-kn} \right ) e^{-j \omega n} & \text{(subs. IDFT)} \end{align*} $$
Here $X_k = X(e^{j\frac{2\pi}{N}k}), \, W_N = e^{-j\frac{2\pi}{N}}$, and let $\omega_k = \frac{2\pi}{N} k$, then rearrange:
$$ \begin{align*} X(e^{j\omega}) = \sum_{k=0}^{N-1} X_k \left ( \frac{1}{N} \sum_{n=0}^{N-1} e^{j(\omega_k - \omega) n} \right ) \end{align*} $$
So the DTFT of an N-sequence is uniquely determined by N uniformly spaced (between $0$ and $2\pi$) frequency samples. The term in the parentheses is an interpolation function (it's actually the digital sinc).
(Check whether your $x_n$ is periodic to 150 samples).
What happens when you zero pad the DFT to length $P > N$ (add $P-N$ more $X_k = 0$ terms), and take the IDFT? You interpolate to a higher sampling rate (since $\omega_k = \frac{2\pi}{P}k$ is less than before).
Note that you're adding points after the first $N/2$ samples, explained here.