Number of ways in which 3 tickets can be selected from a set of 500 tickets numbered 1,2,3..., 500 so that the number on them are in arithmetic progression is 500 (b) `^500 C_3` (c) 62250 (d) None of these

Options

A. 500
B. `.^(500)C_(3)`
C. `3996`
D. None of these

Answer

To solve the problem of selecting 3 tickets from a set of 500 tickets numbered from 1 to 500 such that the numbers on them are in arithmetic progression, we can follow these steps:

### Step 1: Understanding Arithmetic Progression (AP)
In an arithmetic progression, the middle term is the average of the other two terms. If we denote the three selected tickets as A, B, and C, then they must satisfy the condition:
\[ B = \frac{A + C}{2} \]
This implies that \( A + C = 2B \).

### Step 2: Setting the Range for A, B, and C
Since the tickets are numbered from 1 to 500, we can express A, B, and C in terms of their positions in the sequence. Let’s assume:
- A = \( x \)
- B = \( y \)
- C = \( z \)

From the AP condition, we have:
\[ y = \frac{x + z}{2} \]
This means \( x + z = 2y \).

### Step 3: Finding the Conditions for A, B, and C
To ensure that A, B, and C are valid ticket numbers:
1. \( x \) must be less than \( y \) and \( z \) must be greater than \( y \).
2. The values of \( x \), \( y \), and \( z \) must all be within the range of 1 to 500.

### Step 4: Counting the Valid Combinations
To find the number of valid combinations, we can consider the following:
- Let \( d \) be the common difference of the arithmetic progression. Then we can express:
- \( A = y - d \)
- \( B = y \)
- \( C = y + d \)

For \( A \) and \( C \) to be valid ticket numbers:
- \( y - d \geq 1 \) (which gives \( d \leq y - 1 \))
- \( y + d \leq 500 \) (which gives \( d \leq 500 - y \))

Thus, the maximum value of \( d \) is:
\[ d \leq \min(y - 1, 500 - y) \]

### Step 5: Summing Over Possible Values of y
The value of \( y \) can range from 2 to 499 (since \( y \) must be at least 2 to allow for \( A \) and at most 499 to allow for \( C \)). For each valid \( y \):
- The number of valid \( d \) values is given by \( \min(y - 1, 500 - y) \).

### Step 6: Calculating the Total Number of Combinations
We can calculate the total number of combinations by summing the valid \( d \) values for each \( y \):
\[ \text{Total} = \sum_{y=2}^{499} \min(y - 1, 500 - y) \]

### Step 7: Evaluating the Sum
1. For \( y = 2 \) to \( y = 250 \), \( \min(y - 1, 500 - y) = y - 1 \).
2. For \( y = 251 \) to \( y = 499 \), \( \min(y - 1, 500 - y) = 500 - y \).

Calculating these two parts:
- From \( y = 2 \) to \( y = 250 \):
\[ \sum_{y=2}^{250} (y - 1) = \sum_{k=1}^{249} k = \frac{249 \times 250}{2} = 31125 \]
- From \( y = 251 \) to \( y = 499 \):
\[ \sum_{y=251}^{499} (500 - y) = \sum_{k=1}^{249} k = 31125 \]

### Final Calculation
Adding both parts together gives:
\[ 31125 + 31125 = 62250 \]

### Conclusion
Thus, the total number of ways to select 3 tickets such that their numbers are in arithmetic progression is **62250**.