This problem involves determining how long it will take to fill the water tank with an increasing filling rate. Let’s break it down:

Step 1: Determine the rate at which the tank fills.

The amount of water filled increases in an arithmetic sequence. The amounts of water filled in the first few hours are given as:

• 50 liters in the 1st hour,
• 175 liters in the 2nd hour,
• 300 liters in the 3rd hour, and so on.

This is an arithmetic progression (AP) where:

• First term ($a$) = 50 liters,
• Common difference ($d$) = 175 - 50 = 125 liters.

So, the amount of water filled in the $n$-th hour is given by the formula for the $n$-th term of an arithmetic progression: $T_n = a + (n - 1) \cdot d$ Where:

• $T_n$ is the amount filled in the $n$-th hour,
• $a = 50$,
• $d = 125$.

Step 2: Sum of the water filled over time.

To find the total amount of water filled after $n$ hours, we need to sum the terms of the arithmetic progression. The sum of the first $n$ terms of an arithmetic sequence is given by the formula: $S_n = \frac{n}{2} \cdot [2a + (n - 1)d]$ We want this total to equal the capacity of the tank, which is 25,500 liters: $S_n = 25,500$ Substitute $a = 50$, $d = 125$, and solve for $n$: $\frac{n}{2} \cdot [2(50) + (n - 1)(125)] = 25,500$ $\frac{n}{2} \cdot [100 + 125n - 125] = 25,500$ $\frac{n}{2} \cdot (125n - 25) = 25,500$ $n(125n - 25) = 51,000$ $125n^2 - 25n - 51,000 = 0$ This is a quadratic equation. Let's solve it.

Step 3: Solving the quadratic equation.

The quadratic equation is: $125n^2 - 25n - 51,000 = 0$ We can solve this using the quadratic formula: $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 125$, $b = -25$, and $c = -51,000$. Let's solve this next.The positive solution to the quadratic equation is approximately $n \approx 20.3$. Since $n$ represents the number of hours, we round up to the nearest whole number, meaning it will take about 21 hours to fill the tank.

Would you like further details or have any questions about this process?

Here are some related questions you might consider:

1. What if the tank’s capacity were larger? How would it affect the time to fill?
2. How would the time change if the common difference in the filling rate were smaller?
3. What if the rate of filling increased geometrically instead of arithmetically?
4. How much water is filled after 10 hours?
5. Could this method be applied to tanks of irregular shapes?

Tip: When solving quadratic equations, always check for both positive and negative roots to ensure you choose the meaningful solution in context.