Looking the follow schematics you can see two resistors 15k and 27k that are controlling the gate voltage of the PUT (programmable unijunction transistor) transistor. My question is why we need two resistors to control the PUT's Gate, while we can do our job with only one carefully selected resistor?
When the 15K resistor is left out, the gate of the SCR will be pulled down to the negative terminal (which is probably ground, but isn't shown in the schematic). When it is pulled down, the SCR will never turn on.
When the 27K resistor is left out, the gate of the SCR is pulled up. This will (almost) always turn on the SCR (provided that the terminal voltage (6V in case of this schematic) is high enough such that the gate trigger voltage is exceeded and the resistor (here 27K) is such that the gate trigger current is exceeded).
Using a voltage divider only makes sense if the terminal voltage can differ. Otherwise, for a fixed voltage, one can leave out the 27K resistor. For varying terminal voltage, tweaking the voltage divider determines at which terminal voltage the SRC is triggered.
You can use one resistor if you have an appropriate voltage source available. The voltage source needs to be 3.86V and the resistor would be 9.6K to duplicate the behavior of the circuit shown.
The voltage determines the trigger point of the PUT and the resistance determines the valley current. Below from the Onsemi datasheet.
With the normal way PUTs are used, we pick R1/(R1+R2) to program the trigger point (usually something like 2/3 of the supply voltage) and R1*R2/(R1+R2) to program the valley current.
~2/3 or about (1-1/e) of the supply voltage gives you about 1 time constant timing, without getting into the part of the exponential rise where the capacitor voltage rises too slowly and thus becomes much more sensitive to PUT temperature and noise on the supply.