If the sum of the first m terms of an AP is n and the sum of first n terms is m. Then, show that the sum of its first (m+n) terms is -(m+n)

Sure, let's solve this step by step.

Given that the sum of the first m terms of an AP is n and the sum of the first n terms is m.

Let's denote the first term of the AP as 'a' and the common difference as 'd'.

1. The sum of the first m terms of an AP (S_m) is given by the formula: S_m = m/2 * (2a + (m-1)d). According to the problem, S_m = n.

2. Similarly, the sum of the first n terms of an AP (S_n) is given by the formula: S_n = n/2 * (2a + (n-1)d). According to the problem, S_n = m.

3. We need to find the sum of the first (m+n) terms of the AP (S_m+n). The formula for this is: S_m+n = (m+n)/2 * (2a + (m+n-1)d).

4. Substitute the values of S_m and S_n from steps 1 and 2 into the equation from step 3.

5. After substitution, we get: (m+n)/2 * (2a + (m+n-1)d) = m + n.

6. Simplify the equation to get: 2a + (m+n-1)d = 2.

7. Further simplification gives: a + (m+n-1)d/2 = 1.

8. But we know from the problem that a + (m-1)d/2 = n/m and a + (n-1)d/2 = m/n.

9. Equating these, we get: n/m = m/n = 1.

10. This implies that m = n.

11. Substituting m = n in the equation from step 7, we get: a + (2m-1)d/2 = 1.

12. This simplifies to: a + md - d/2 = 1.

13. But we know from the problem that a + md = n and a + nd = m.

14. Substituting these values, we get: n - d/2 = 1 and m - d/2 = 1.

15. Solving these equations, we get: d