The number of ways in which three numbers in arithmetic progression can be selected from `{1, 2, 3, ………., 50}` is

Options

A. 276
B. 600
C. 840
D. 640

Answer

To solve the problem of finding the number of ways to select three numbers in arithmetic progression from the set {1, 2, 3, …, 50}, we can follow these steps:

### Step 1: Understanding Arithmetic Progression (AP)
Three numbers \( A, B, C \) are in arithmetic progression if the middle number \( B \) is the average of \( A \) and \( C \). This can be expressed mathematically as:
\[ 2B = A + C \]
This implies that \( A \) and \( C \) must be chosen such that \( B \) is an integer.

### Step 2: Expressing the Numbers
Let’s denote the three numbers in AP as:
- \( A = B - d \)
- \( B = B \)
- \( C = B + d \)

Here, \( d \) is the common difference. The numbers \( A \), \( B \), and \( C \) must all lie within the range of 1 to 50.

### Step 3: Constraints on \( B \) and \( d \)
For \( A \) and \( C \) to be valid:
- \( A = B - d \geq 1 \) implies \( B - d \geq 1 \) or \( d \leq B - 1 \)
- \( C = B + d \leq 50 \) implies \( B + d \leq 50 \) or \( d \leq 50 - B \)

Thus, we have:
\[ d \leq \min(B - 1, 50 - B) \]

### Step 4: Counting Valid \( d \) for Each \( B \)
For each value of \( B \) from 1 to 50, we need to count how many valid \( d \) values exist. The maximum value of \( d \) is given by:
\[ d_{\text{max}} = \min(B - 1, 50 - B) \]
The number of valid \( d \) values for each \( B \) is:
\[ d_{\text{max}} = \min(B - 1, 50 - B) \]

### Step 5: Summing Over All Possible \( B \)
Now, we will sum the valid \( d \) values for each \( B \):
- For \( B = 1 \): \( d_{\text{max}} = 0 \) (no valid \( d \))
- For \( B = 2 \): \( d_{\text{max}} = 1 \) (valid \( d = 1 \))
- For \( B = 3 \): \( d_{\text{max}} = 2 \) (valid \( d = 1, 2 \))
- ...
- For \( B = 25 \): \( d_{\text{max}} = 24 \) (valid \( d = 1, 2, ..., 24 \))
- For \( B = 26 \): \( d_{\text{max}} = 24 \) (valid \( d = 1, 2, ..., 24 \))
- ...
- For \( B = 49 \): \( d_{\text{max}} = 1 \) (valid \( d = 1 \))
- For \( B = 50 \): \( d_{\text{max}} = 0 \) (no valid \( d \))

### Step 6: Calculating Total Combinations
The total number of valid combinations is the sum of valid \( d \) values for each \( B \):
\[ \text{Total combinations} = \sum_{B=2}^{25} (B - 1) + \sum_{B=26}^{49} (50 - B) \]

Calculating these sums:
- From \( B = 2 \) to \( B = 25 \):
\[ \sum_{k=1}^{24} k = \frac{24 \times 25}{2} = 300 \]
- From \( B = 26 \) to \( B = 49 \):
\[ \sum_{k=1}^{24} k = 300 \]

Thus, the total number of ways to select three numbers in arithmetic progression from the set {1, 2, ..., 50} is:
\[ 300 + 300 = 600 \]

### Final Answer
The number of ways in which three numbers in arithmetic progression can be selected from {1, 2, 3, …, 50} is **600**.

---