Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?
Step-by-Step Solution
Key Concept: For two arithmetic progressions with the same common difference \(d\), the difference between their \(n^{th}\) terms is independent of \(n\) and equals the difference between their first terms: \(T_{n}^{(1)}-T_{n}^{(2)} = a_1-a_2\).
1. Let the first AP be \(a_1, a_1+d, a_1+2d,\dots\) and the second AP be \(a_2, a_2+d, a_2+2d,\dots\).\
2. General term of an AP: \(T_n = a + (n-1)d\).\
3. 100th term of the first AP: \(T_{100}^{(1)} = a_1 + 99d\).\
100th term of the second AP: \(T_{100}^{(2)} = a_2 + 99d\).\
4. Given \(T_{100}^{(1)} - T_{100}^{(2)} = 100\):\
\[ (a_1 + 99d) - (a_2 + 99d) = a_1 - a_2 = 100. \]
Hence, \(a_1 - a_2 = 100\).\
5. 1000th term of the first AP: \(T_{1000}^{(1)} = a_1 + 999d\).\
1000th term of the second AP: \(T_{1000}^{(2)} = a_2 + 999d\).\
6. Difference between the 1000th terms:\
\[ T_{1000}^{(1)} - T_{1000}^{(2)} = (a_1 + 999d) - (a_2 + 999d) = a_1 - a_2. \]
Using the result from step 4, \(a_1 - a_2 = 100\).\
7. Therefore, the difference between the 1000th terms is also \(100\).
Correct Answer: 100