**Q) If the sum of first m terms of an A.P. is same as sum of its first n terms (m ≠ n), then show that the sum of its first (m + n) terms is zero.**

**Ans: **

**Step 1: **We know that the sum of first m terms of an AP is given by:

S_{m} = (2 a + (m – 1) d)

Similarly, Sum of first n terms: S_{n} = (2 a + (n – 1) d)

By given condition, S_{m} = S_{n}

∴ (2 a + (m – 1) d) = (2 a + (n – 1) d)

∴ m (2 a + (m – 1) d) = n (2 a + (n – 1) d)

∴ 2 a m + m d (m – 1) = 2 a n + n d (n – 1)

∴ 2 a m + d m ^{2 } – m d = 2 a n + d n ^{2 } – n d

∴ 2 a m – 2 a n + d m ^{2 } – d n ^{2 } – m d + n d = 0

∴ 2 a (m – n ) + d (m ^{2 } – n ^{2 }) – d (m – n) = 0

∴ 2 a (m – n ) + d (m – n) (m + n) – d (m – n) = 0

∴ (m – n ) (2 a + d (m + n) – d) = 0

∴ 2 a + (m + n – 1) d = 0 …………. (i)

**Step 2: **Sum of first (m + n) terms: S_{(m + n)} = (2 a + (m + n – 1) d)

substituting the value of (m + n – 1) d from equation (i), we get:

S_{(m + n)} = (0) = 0

**Therefore, sum of first (m + n) terms is zero.**

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