Answer
The current through $R_1$ is less than it was previously.
Work Step by Step
Let $\Delta V$ be the potential difference across the battery.
Initially, the potential difference across $R_1$ is $\Delta V$
After $R_2$ is added in series, the sum of the potential differences across $R_1$ and $R_2$ is $\Delta V$
Therefore, the potential difference across $R_1$ is less than it was previously.
Initially, the current through $R_1$ is $\frac{\Delta V}{R_1}$
After $R_2$ is added in series, the potential difference across $R_1$ decreases, so the current through $R_1$ also decreases.
The current through $R_1$ is less than it was previously.
