To find 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: Understand the condition for arithmetic progression
Three numbers \( x_1, x_2, x_3 \) are in arithmetic progression if:
\[
2x_2 = x_1 + x_3
\]
This means that \( x_2 \) is the average of \( x_1 \) and \( x_3 \).
### Step 2: Express the numbers in terms of a common difference
Let’s denote the first term of the arithmetic progression as \( a \) and the common difference as \( d \). Then we can express the three terms as:
\[
x_1 = a, \quad x_2 = a + d, \quad x_3 = a + 2d
\]
Here, \( a \) must be chosen such that \( x_3 \leq 50 \).
### Step 3: Determine the range for \( a \) and \( d \)
From the expression \( x_3 = a + 2d \leq 50 \), we can derive:
\[
a + 2d \leq 50 \implies 2d \leq 50 - a \implies d \leq \frac{50 - a}{2}
\]
Thus, for each valid \( a \), \( d \) can take values from \( 1 \) to \( \frac{50 - a}{2} \).
### Step 4: Count the valid pairs \( (a, d) \)
Now, we need to find the number of valid pairs \( (a, d) \):
- The smallest value for \( a \) is \( 1 \).
- The largest value for \( a \) is \( 50 \), but since \( d \) must be at least \( 1 \), \( a \) can go up to \( 48 \) (because if \( a = 49 \), \( d \) can only be \( 0.5 \), which is not allowed).
Thus, \( a \) can take values from \( 1 \) to \( 48 \).
### Step 5: Calculate the total number of ways
For each value of \( a \):
- If \( a = 1 \), \( d \) can be \( 1, 2, \ldots, 24 \) (24 options).
- If \( a = 2 \), \( d \) can be \( 1, 2, \ldots, 24 \) (24 options).
- Continuing this way, we see that as \( a \) increases, the maximum value of \( d \) decreases.
The number of valid \( d \) values for each \( a \) is:
- For \( a = 1 \) to \( 24 \): \( 24, 23, 22, \ldots, 1 \) (which gives us \( 24 \) options).
- For \( a = 25 \) to \( 48 \): \( 12, 11, 10, \ldots, 1 \) (which gives us \( 12 \) options).
### Step 6: Sum the valid combinations
The total number of ways can be calculated as:
\[
\text{Total} = 24 + 23 + 22 + \ldots + 1 + 12 + 11 + \ldots + 1
\]
Using the formula for the sum of the first \( n \) natural numbers:
\[
\text{Sum} = \frac{n(n + 1)}{2}
\]
For \( n = 24 \):
\[
\text{Sum} = \frac{24 \cdot 25}{2} = 300
\]
For \( n = 12 \):
\[
\text{Sum} = \frac{12 \cdot 13}{2} = 78
\]
Thus, the total number of ways is:
\[
300 + 78 = 378
\]
### Final Answer
The number of ways in which three numbers in arithmetic progression can be selected from the set {1, 2, 3, ..., 50} is **378**.
