Lower duty cycle less efficient?
In this specific case, the power dissipated by the resistor would double.
The more general principal, which is always applicable in all cases is that the average power dissipated by a resistor is the mean value of the instantaneous power calculated over one period.
So even if you have an irregular or arbitrary repeating voltage or current waveform, you can calculate the average power using the mean value of the instantaneous power. Instantaneous power can be calculated using V(t)^2/R or I(t)^2 * R or V(t) * I(t). Obviously you must calculate the mean over one period (or any whole number of periods). You can't calculate mean power over a fraction of a period.