Let T_r be the r th term of an A.P. If mT_m = nT _n, then show that T_m+n = 0

Given:     r^th term of A. P. = Tr

Then,   m^th term of A. P. = Tm

and      n^th term of A. P. = Tn

According to question –

mT_m = nT_n

⇒ m [a + (m – 1) d] = n [a + (n – 1) d]

Where a is first term and d is the common difference of given A.P.

Hence Proved