Amount of flour for n-th recipe test:
Note that this is the term of an arithmetic progression, not a geometric progression. A GP uses a constant ratio to move from one term to the next, whereas an AP uses a constant additive to move from one term to the next. In this case, 20g of flour is the additive for each recipe trial.
Total amount of flour for N recipe tests, where 1≤n≤N:
When N=10, F(10)=2000+900=2.9kg.
For the first two trials the amount of sugar used would be 100g and 110g, based on using half as much sugar as flour, totalling 0.21kg, leaving 1.5-0.21=1.29kg of sugar. (And she will use 0.42kg of flour in two trials.) The amount of sugar for the remaining trials must not exceed 1.29kg.
On the third trial the amount of flour is 240g, requiring 120g (half the flour amount) of sugar. The amount of flour increases by 20g for subsequent trials, so the sugar will increase by 10g. So the formula for sugar over N₁ trials is:
S(N₁)=120(N₁)+10∑(n-1) where 1≤n≤N₁. The total number of trials is N=N₁+2.
There is something of an ambiguity here in the question regarding the initial amount of sugar for the third trial: it could be 100g or 120g. Let’s look at these separately: first, 120g:
Total amount of sugar:
S(N₁)=120N₁+5N₁(N₁-1) based on an initial 120g of sugar (on the third trial) increasing by 10g for the remaining trials until the sugar supply is depleted.
S(N₁)≤1290 so we need to find N₁:
Solving N²+23N₁-258=0 using the quadratic formula:
(Note that the other zero is negative and has no meaning in this context.)
N must be an integer so N₁≤8.25, making N₁=8 and N=10, the total number of trials.
This corresponds to 960+280=1.24kg of sugar. Plus the amount of sugar used in the first two trials: 1.24+0.21=1.45kg out of 1.5kg.
Now we use 100g as the initial amount of sugar:
N₁=(-19+√(361+1032))/2=9.16 approx, so integer N₁=9 and N=11, the total number of trials.