Number of A.P that can be formed from the set of numbers `{1,2,3,…..,1000}` such that first term of A.p is always . 1 and last term is 1000 _______

Answer

To find the number of arithmetic progressions (A.P.) that can be formed from the set of numbers {1, 2, 3, ..., 1000} such that the first term is always 1 and the last term is 1000, we can follow these steps:

### Step 1: Understand the A.P. Formula
The nth term of an arithmetic progression can be expressed as:
\[ T_n = a + (n - 1) \cdot d \]
where:
- \( T_n \) is the nth term,
- \( a \) is the first term,
- \( d \) is the common difference,
- \( n \) is the number of terms.

### Step 2: Set Up the Equation
Given that the first term \( a = 1 \) and the last term \( T_n = 1000 \), we can set up the equation:
\[ 1000 = 1 + (n - 1) \cdot d \]

### Step 3: Simplify the Equation
Rearranging the equation gives:
\[ 999 = (n - 1) \cdot d \]
This means:
\[ n - 1 = \frac{999}{d} \]
Thus, we can express \( n \) as:
\[ n = \frac{999}{d} + 1 \]

### Step 4: Determine the Factors of 999
To find the number of possible values for \( d \), we need to find the factors of 999. First, we factor 999:
\[ 999 = 3^3 \cdot 37^1 \]

### Step 5: Calculate the Number of Factors
Using the formula for finding the number of factors from the prime factorization:
If \( n = p_1^{k_1} \cdot p_2^{k_2} \cdots p_m^{k_m} \), then the number of factors is given by:
\[ (k_1 + 1)(k_2 + 1) \cdots (k_m + 1) \]

For \( 999 = 3^3 \cdot 37^1 \):
- The exponent of 3 is 3, so \( k_1 + 1 = 3 + 1 = 4 \).
- The exponent of 37 is 1, so \( k_2 + 1 = 1 + 1 = 2 \).

Thus, the total number of factors is:
\[ 4 \cdot 2 = 8 \]

### Step 6: Conclusion
Each factor of 999 corresponds to a different value of \( d \), and since \( d \) can take on 8 different values, we conclude that the number of arithmetic progressions that can be formed is:
**8**