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If m times the m^{th} term of an A.P. is equal to n times n^{th} term then show that the (m + n)^{th} term of the A.P. is zero.

**Sol**. Let a be the first term and be the common difference.

∴ mt_{m} = nt_{n} [Given]

∴ m[a + (m – 1)d] = n[a + (n – 1)d]

∴ am + m^{2}d – md = an + n^{2}d – nd

∴ am – an + m^{2}d – n^{2}d – md + nd = 0

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

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

∴ (m – n) [a + d(m + n) – d] = 0 [Taking (m – n) common out of all terms]

∴ (m – n) [a + (m + n – 1)d] = 0

∴ m – n = 0 or a + (m + n – 1)d = 0

t_{m} and t_{n} are two different terms

∴ m ≠ n ∴ m – n ≠ 0

∴ a + (m + n – 1)d = 0 ...(I)

Now, t(m+n) = a + (m + n – 1)d ...(II)

∴ **t(m+n) = 0** [From (I) and (II)]