If m times the mth term of an A.P. is equal to n times its nth term, then show that the (m + n)th term of the A.P. is zero.
We know that nth term of A.P. , tn = a + (n – 1) d.
⇒ First, tn = a + (n – 1) d
⇒ Then, tm = a + (m – 1) d
Given, mtm = ntn
⇒ m [a + (m – 1) d] = n [a + (n – 1) d]
⇒ ma + m2d – md = na + n2d – nd
⇒ ma – na + m2d – n2d – md + nd = 0
⇒ a (m – n) + d (m2 – n2) – d (m – n) = 0
We know that a2 – b2 = (a – b) (a + b)
⇒ (m – n) [a + d (m + n) – d] = 0
⇒ [a + d (m + n) – d] = 0
⇒ a + (m + n – 1) d = 0
∴ (m + n)th term, tm + n = 0
Hence proved.
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