Q29 of 47 Page 1

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

Given that Sm = Sn

We have to show that S(m + n) = 0.


We know that the sum of the first n terms of an AP is given by Sn = [n (2a + (n – 1) d)] / 2


∴ [m (2a + (m – 1) d)] / 2 = [n (2a + (n - 1) d] / 2


⇒ [2am + m (m – 1) d] = [2an + n (n – 1) d] / 2


⇒ 2a (m – n) + [(m2 – n2) – (m – n)] d = 0


⇒ (m – n) [2a + (m + n – 1)] d = 0


∴ [2a + (m + n – 1)] d = 0 … (1)


Now,


S(m + n) = [(m + n) (2a + (m + n – 1)) d] / 2


From (1), we get


⇒ S(m + n) = [(m + n) 0] / 2


∴ S(m + n) = 0


Ans. From the given conditions, we showed that S(m + n) = 0.


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