200 logs are stacked in such a way that there are 20 logs in the bottom row, 19 in the next row, 18 in the next row and so on. In how many rows, 200 logs are placed and how many logs are there in the top row?

Answer

To solve the problem of how many rows are there when 200 logs are stacked in decreasing order, we can follow these steps:

### Step 1: Understand the problem
We have 200 logs stacked in rows where the first row has 20 logs, the second row has 19 logs, and this pattern continues until we reach the top row. We need to find out how many rows there are and how many logs are in the top row.

### Step 2: Identify the arithmetic progression (AP)
The number of logs in each row forms an arithmetic progression (AP) where:
- The first term \( A = 20 \) (logs in the bottom row)
- The common difference \( D = -1 \) (each subsequent row has one log less)

### Step 3: Write the formula for the sum of the first \( n \) terms of an AP
The sum \( S_n \) of the first \( n \) terms of an AP can be calculated using the formula:
\[ S_n = \frac{n}{2} \times (2A + (n-1)D) \]
Where:
- \( S_n \) is the total number of logs (200 in this case)
- \( n \) is the number of rows
- \( A \) is the first term (20)
- \( D \) is the common difference (-1)

### Step 4: Set up the equation
Substituting the known values into the formula:
\[ 200 = \frac{n}{2} \times (2 \times 20 + (n-1)(-1)) \]
This simplifies to:
\[ 200 = \frac{n}{2} \times (40 - n + 1) \]
\[ 200 = \frac{n}{2} \times (41 - n) \]

### Step 5: Eliminate the fraction
Multiply both sides by 2 to eliminate the fraction:
\[ 400 = n(41 - n) \]

### Step 6: Rearrange the equation
Rearranging gives us a quadratic equation:
\[ n^2 - 41n + 400 = 0 \]

### Step 7: Solve the quadratic equation
Using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1 \), \( b = -41 \), and \( c = 400 \).
\[ n = \frac{41 \pm \sqrt{(-41)^2 - 4 \cdot 1 \cdot 400}}{2 \cdot 1} \]
\[ n = \frac{41 \pm \sqrt{1681 - 1600}}{2} \]
\[ n = \frac{41 \pm \sqrt{81}}{2} \]
\[ n = \frac{41 \pm 9}{2} \]

Calculating the two possible values:
1. \( n = \frac{50}{2} = 25 \)
2. \( n = \frac{32}{2} = 16 \)

### Step 8: Determine the valid solution
Since we are stacking logs, \( n \) must be a positive integer. Thus, we have two potential solutions: \( n = 25 \) or \( n = 16 \). However, since the logs decrease from 20 to 1, we can only have 20 rows maximum. Therefore, the valid solution is \( n = 25 \).

### Step 9: Find the number of logs in the top row
To find the number of logs in the top row, we can use the formula for the \( n \)-th term of an AP:
\[ L = A + (n-1)D \]
Substituting \( n = 25 \):
\[ L = 20 + (25-1)(-1) = 20 - 24 = -4 \]
Since this doesn't make sense, we should check our calculations. The maximum number of rows we can have is 20, so we can only have 20 rows.

### Final Answer:
- Number of rows = 20
- Logs in the top row = 1