Two A.P .s have the same common difference . The difference between their `100^(th)` term is 100, what is the difference between their `1000^(th)` terms ?

Answer

To solve the problem step by step, we need to understand the properties of Arithmetic Progressions (APs).

### Step-by-Step Solution:

1. **Define the Terms of the A.P.s**:
Let the first A.P. be defined as:
- First term: \( a_1 \)
- Common difference: \( d \)
- \( n^{th} \) term formula: \( T_n = a_1 + (n-1)d \)

Let the second A.P. be defined as:
- First term: \( a_2 \)
- Common difference: \( d \)
- \( n^{th} \) term formula: \( T_n' = a_2 + (n-1)d \)

2. **Find the 100th Terms**:
The 100th term of the first A.P. is:
\[ T_{100} = a_1 + 99d \]
The 100th term of the second A.P. is:
\[ T_{100}' = a_2 + 99d \]

3. **Set Up the Difference Between the 100th Terms**:
According to the problem, the difference between the 100th terms of the two A.P.s is given as:
\[ T_{100} - T_{100}' = 100 \]
Substituting the expressions we found:
\[ (a_1 + 99d) - (a_2 + 99d) = 100 \]
Simplifying this gives:
\[ a_1 - a_2 = 100 \]

4. **Find the 1000th Terms**:
The 1000th term of the first A.P. is:
\[ T_{1000} = a_1 + 999d \]
The 1000th term of the second A.P. is:
\[ T_{1000}' = a_2 + 999d \]

5. **Set Up the Difference Between the 1000th Terms**:
The difference between the 1000th terms is:
\[ T_{1000} - T_{1000}' = (a_1 + 999d) - (a_2 + 999d) \]
Simplifying this gives:
\[ a_1 - a_2 = 100 \]

6. **Conclusion**:
Since we already established that \( a_1 - a_2 = 100 \), the difference between the 1000th terms is also:
\[ T_{1000} - T_{1000}' = 100 \]

### Final Answer:
The difference between their 1000th terms is **100**.