Considering it looked like a geometric series to me (without doing any computations, just a glance) I suspect he means exponential decay. The two (decay and growth) are basically the same thing, so he’s not wrong, but k < 1

Something like adding x lbs weight requires y lbs fuel, where y is x/2 or something, means you’re adding x + x/2 + x/4 + x/8 + … which is exponential (look at the denominator). But I would never try to solve it that way. It’s a simple system of equations, I think.

I suspect this will match an earlier post but lets just do this anyway:

total_weight = fuel_weight + initial_weight

fuel_weight = total_weight*gallons_per_pound*weight_per_gallon

I tried to avoid ambiguity but gallons_per_pound (GPP) is the amount of gallons required to lift 1 pound, and weight_per_gallon (WPG) is the actual weight of a gallon of fuel. The logic here is that we *know* the total weight is the initial+the fuel, and we also know that the weight of the fuel is capable of lifting the total weight, so all I did was back convert. Now, solving the system…

total_weight = total_weight*GPP*WPG + initial_weight

total_weight - total_weight*GPP*WPG = initial_weight

total_weight*(1-GPP*WPG) = initial weight

total_weight = initial_weight/(1-GPP*WPG)

In the problem given, WPG =0.8, GPP = 0.3125 and initial_weight = 16000kg thus total_weight = 21333.3 kg, fuel_weight = 5333.3 kg and the number of liters is 6666.7L.

The answer, in one equation then, would be **total_weight = initial_weight/(1-liters_to_lift_a_kilogram*kilograms_per_liter)**

—Edit: I should have just simplified this to total_weight = initial_weight/0.75, as the long names are really constants provided in the initial problem!—

Note: I was halfway through the post when I realized that you actually *weren’t* using crappy imperial units (great irony, as I’m American and they make no sense to me). But I didn’t change it because the 3 letter abbreviation is easier in imperial units.

Note: The 1-x term scared me (what if x>1?) but it makes sense. If it takes more than a kilogram of fuel to lift a kilogram, you’re going to have negative values!

Note: This is only a solution to the problem as given, and doesn’t take into account the compounding effects of, ultimately, F = MA (where burned fuel is generating a relatively constant F, and A depends on the M of the plane). The formulas given by a previous poster is a MUCH more accurate model of reality.

Note: I suck at math, and probably screwed this up.