That's a very puzzling piece of code. Why are you solving for Qa when you already have Qa = (100000 - 200 Pa/3) (1.01)^(i - 1)? In order to get rid of the Pa, you're going to have to assign its value somewhere above the bit of code you've given. Here's a way to find out if Mathematica knows the value of a variable:
Type in the variable in a new cell, and hit Shift-Enter (that is, evaluate the cell). If Mathematica spits back the variable right back at you, then it hasn't been assigned a value. But if it returns something other than the variable name, then it has been assigned.
Also, I don't think this code is going to do what you want: you've assigned Qa using the = assignment operator. That's a one-shot deal. Whatever is in the for loop initiation gets assigned once. So, you're going to have
Qa=(100000 - 200 Pa/3) (1.01)^(1 - 1) = 100000 - 200 Pa/3, no matter what value of i you loop through. Try the delayed definition of :=. Then, whenever you change the variable i, Qa will change right along with you. Another way to handle it is to put the Qa assignment in the main body of the for loop, instead of in the initialization part.
Hope this helps.