How did you convert your 3645÷10950 into fraction

Both the numerator and the denominator are divisible by 5 (since they end in 5 and 0, respectively).

$$\frac{3645 \div 5}{10950 \div 5} = \frac{729}{2190}$$

The sum of the digits of 729 ($7+2+9=18$) is divisible by 3.
The sum of the digits of 2190 ($2+1+9+0=12$) is divisible by 3.
So, both numbers are divisible by 3.

$$\frac{729 \div 3}{2190 \div 3} = \frac{243}{730}$$

The prime factorization of 243 is $3^5$.
The prime factorization of 730 is $2 \times 5 \times 73$.
Since there are no common prime factors, the fraction $\frac{243}{730}$ is in its simplest form.

Therefore, $\frac{3645}{10950} = \boxed{\frac{243}{730}}$.

Clarification on the common ratio used in the previous solution:
In the previous solution, it was stated that the common ratio $r = \frac{3,645}{10,950} = \frac{1}{3}$. This was an error in calculation. As shown above, $\frac{3,645}{10,950} = \frac{243}{730}$.

However, if we calculate the ratio between the third and second prize amounts:

$$r' = \frac{1,215}{3,645}$$ $$r' = \frac{1,215 \div 1,215}{3,645 \div 1,215} = \frac{1}{3}$$

This ratio is exactly $\frac{1}{3}$.

The problem statement implies a geometric progression with the phrase "as so on". Given that the ratio between the second and third terms is exactly $\frac{1}{3}$, it was assumed in the previous solution that the intended common ratio for the entire progression was $\frac{1}{3}$, despite the first term not perfectly fitting this ratio ($10,950 \times \frac{1}{3} = 3,650$, which is not 3,645). The previous solution proceeded with $a_1 = 10,950$ and $r = \frac{1}{3}$ to answer the questions, interpreting the problem as having an intended common ratio of $1/3$.