If in an A.P., $T_{m} = n, T_{n} = m$ , prove $d = - 1$

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Solution

$a + (m - 1) d = n$ $(1)$ $a + (n - 1) d = m$ _ $(2)$
$(1) - (2)$
$a + (m - 1) d - a - (n - 1) d = n - m$
$((m - 1) - (n - 1)) d = n - m$
$(m - 1 - n + 1) d = n - m$
$d = \frac{n - m}{m - n} = \frac{- (m - n)}{(m - n)}$
$d = - 1$

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