Show that the number of ways in which three numbers in arithmetical progresssion can be selected from 1,2,3,……..n is `1/4(n-1)^(2)` or `1/4n(n-2)` according as n is odd or even.

Answer

To solve the problem of finding the number of ways to select three numbers in arithmetic progression from the set {1, 2, 3, ..., n}, we will break it down step by step.

### Step 1: Understand the Arithmetic Progression (AP)
Let the three terms in arithmetic progression be represented as:
- First term: \( A \)
- Second term: \( A + D \)
- Third term: \( A + 2D \)

where \( A \) is the first term and \( D \) is the common difference.

### Step 2: Set the Constraints
The terms must satisfy the following conditions:
1. \( A \) must be a natural number (i.e., \( A \geq 1 \)).
2. The last term \( A + 2D \) must not exceed \( n \) (i.e., \( A + 2D \leq n \)).

From the second condition, we can derive:
\[ A \leq n - 2D \]

### Step 3: Determine the Range for \( D \)
Since \( A \) must be at least 1, we also have:
\[ 1 \leq A \leq n - 2D \]

This gives us:
\[ D \leq \frac{n - 1}{2} \]

### Step 4: Count the Possible Values of \( A \)
For each fixed value of \( D \), the possible values of \( A \) range from 1 to \( n - 2D \). Therefore, the number of possible values for \( A \) is:
\[ n - 2D \]

### Step 5: Calculate the Total Number of Combinations
Now, we need to sum the possible values of \( A \) for each valid \( D \):
- The values of \( D \) can range from 1 to \( \left\lfloor \frac{n - 1}{2} \right\rfloor \).

Thus, the total number of ways to select three numbers in arithmetic progression is given by:
\[ \sum_{D=1}^{\left\lfloor \frac{n - 1}{2} \right\rfloor} (n - 2D) \]

### Step 6: Evaluate the Sum
1. If \( n \) is even:
- Let \( n = 2k \). Then \( D \) can take values from 1 to \( k - 1 \).
- The sum becomes:
\[ \sum_{D=1}^{k-1} (2k - 2D) = 2k(k - 1) - 2\sum_{D=1}^{k-1} D \]
- The sum \( \sum_{D=1}^{k-1} D = \frac{(k-1)k}{2} \).
- Therefore, the total becomes:
\[ 2k(k - 1) - 2 \cdot \frac{(k-1)k}{2} = k(k - 1) = \frac{n(n - 2)}{4} \]

2. If \( n \) is odd:
- Let \( n = 2k + 1 \). Then \( D \) can take values from 1 to \( k \).
- The sum becomes:
\[ \sum_{D=1}^{k} (2k + 1 - 2D) = (2k + 1)k - 2\sum_{D=1}^{k} D \]
- The sum \( \sum_{D=1}^{k} D = \frac{k(k + 1)}{2} \).
- Therefore, the total becomes:
\[ (2k + 1)k - 2 \cdot \frac{k(k + 1)}{2} = k^2 = \frac{(n - 1)^2}{4} \]

### Conclusion
Thus, we have shown that:
- If \( n \) is even, the number of ways is \( \frac{1}{4} n(n - 2) \).
- If \( n \) is odd, the number of ways is \( \frac{1}{4} (n - 1)^2 \).