If the sum of first $m$ terms of an A.P. is same as sum of its first $n$ terms ($m \neq n$), then show that the sum of its first $(m + n)$ terms is zero.
Generated by claude-sonnet-4-6 · 2026-06-15 10:31 · grounding rag
Model Answer
Given: $S_m = S_n$, where $m \neq n$
Using the formula $S_n = \dfrac{n}{2}[2a + (n-1)d]$:
$$\frac{m}{2}[2a + (m-1)d] = \frac{n}{2}[2a + (n-1)d]$$
$$m[2a + (m-1)d] - n[2a + (n-1)d] = 0$$
$$2a(m-n) + d[m(m-1) - n(n-1)] = 0$$
$$2a(m-n) + d[(m^2 - n^2) - (m - n)] = 0$$
$$2a(m-n) + d(m-n)[(m+n) - 1] = 0$$
Since $m \neq n$, divide by $(m - n)$:
$$2a + (m+n-1)d = 0 \quad \cdots (1)$$
Now, sum of first $(m+n)$ terms:
$$S_{m+n} = \frac{(m+n)}{2}[2a + (m+n-1)d] = \frac{(m+n)}{2} \times 0 = \boxed{0}$$
[From equation (1)]
Hence proved.
Source: Chapter 5, Section 5.4 (Sum of first $n$ terms of an AP)
---
Explanation
- The key step is subtracting the two equal sum expressions and factoring out $(m-n)$, which is non-zero, giving $2a + (m+n-1)d = 0$.
- Examiners award marks for: setting up $S_m = S_n$ correctly (1 mark), reaching $2a+(m+n-1)d=0$ (1 mark), and concluding $S_{m+n}=0$ (1 mark).
- Notice that $2a + (m+n-1)d$ is exactly the bracket in the formula for $S_{m+n}$—that's the elegant connection to spot.