Answer
If they are in series, the effective resistance will be closer in value to that of the large resistance because$R_{s}=R_{1}+R_{2}$.
If $R_{1}>>R_{2}$, then $R_{s}\approx R_{1}$.
If they are in parallel, the effective resistance will be closer in value to that of the small resistance because $R_{p}=\frac{R_{1}R_{2}}{R_{1}+R_{2}}$.
If $R_{1}>>R_{2}$, then $R_{p}\approx\frac{R_{1}R_{2}}{R_{1}}=R_{2}$.
Work Step by Step
If they are in series, the effective resistance will be closer in value to that of the large resistance because$R_{s}=R_{1}+R_{2}$.
If $R_{1}>>R_{2}$, then $R_{s}\approx R_{1}$.
If they are in parallel, the effective resistance will be closer in value to that of the small resistance because $R_{p}=\frac{R_{1}R_{2}}{R_{1}+R_{2}}$.
If $R_{1}>>R_{2}$, then $R_{p}\approx\frac{R_{1}R_{2}}{R_{1}}=R_{2}$.