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:

Sm = \frac{m}{2} (2 a + (m – 1) d)

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

By given condition, Sm = Sn

\frac{m}{2} (2 a + (m – 1) d) = \frac{n}{2} (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) = \frac{m + n}{2} (2 a + (m + n – 1) d)

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

S(m + n) = \frac{m + n}{2} (0) = 0

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

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