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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.

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Given, mtm = ntn 

m[a + (m – 1) d] = n [a + (n – 1) d] 

[we know that tn = a + (n – 1)d]

m[a + md – d] = n[a + nd – d] 

ma + m2d – md = na + n2d – nd 

ma – na + m2d – n2d = md – nd

a (m – n) + d (m2 – n2) = d(m – n) 

a (m – n) + d(m + n)(m – n) = d(m – n) ÷ by (m – n) on both sides

a + d (m + n) = d 

a + d(m + n) – d = 0 

a + d(m + n – 1) = 0 … (1) 

To prove, tm + n = 0 

tm + n = a + (m + n – 1)d 

tm + n = 0 (from(1)) 

Hence it is proved.

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