Depending on your goal, and why you placed the capacitor in series with the voltage divider, if your desire is to minimize the capacitive reactance effect of the capacitor on the resistive voltage divider, then you will require as large a capacitance as possible. Allowing |Xc| = 1 / (2 π f C) to approach zero by arbitrarily increasing C without bound would accomplish that goal, but you must also consider the effect of whatever source impedance is driving the series combination of the capacitor and the two resistors, as well as whatever load impedance is in parallel with the lower divider resistor. All this will involve the arithmetic of complex numbers, which clearly you do not seem to understand yet because of your incorrect statement:
Reactance is not the same as resistance, and you cannot simply add 20 kΩ capacitive reactance to 10 kΩ ohmic resistance to obtain 30 kΩ to use in the DC voltage divider equation. The AC voltage divider equation is Vout = Vin [ Zlower / ( Zlower + Zupper ) ]. Impedances are complex numbers and their arithmetic is different than the arithmetic used with real numbers.
A capacitor in series with a resistor has an impedance, |Zeffective| = [Xc2 + R2]1/2. The voltage divider output is NOT Vout = Vin [ Rlower / (Rlower + Rupper + Xc) ]. Instead Vout = Vin [ Rlower / ( Rlower + Zeffective ) ]. The addition of Rlower and Zeffective in the denominator is a complex number addition, and the fraction Rlower / (Rupper + Zeffective) is a complex number.
I will ask again one final time: WTF are you trying to DO?