In an AC series circuit, $R=10\Omega$, $X_L=10\Omega$ and $X_C=20\Omega$. Then, choose the correct options from the following,

Resonance frequency  ${\omega }_0=\frac{1}{\sqrt{LC}}$

From the given values, 

$X_L<X_C \Rightarrow \omega L<\frac{1}{\omega C} \Rightarrow \omega <\frac{1}{\sqrt{LC}} \Rightarrow \omega <{\omega }_0$

$\begin{array}{rcl}\text{As } X_L& =& \omega L \Rightarrow X_L\propto \omega \\ \text{and} X_C& =& \frac{1}{\omega C} \Rightarrow X_C\propto \frac{1}{\omega }\end{array}$

If frequency is decreased from the given value, $X_c$ will further increase. So,( $X_C-X_L$) will increase. Hence, net impedance will increase.

If frequency is decreased from the given value, then Z will increase and hence current will decrease .