In an A.P. if m^th term is n and n^th term is m, show that its r^th term is (m + n - r).

Let first term be a and common difference be d.

Given,

⇒ a_m = n

⇒ a + (m - 1)d = n

⇒ a + md - d = n

⇒ a = n - md + d …….(i)

Also,

⇒ a_n = m

⇒ a + (n - 1)d = m

⇒ a + nd - d = m

Substituting value of a from (i) in above equation,

⇒ n - md + d + nd - d = m

⇒ nd - md = m - n

⇒ d(n - m) = m - n

⇒ d = $\dfrac{m - n}{n - m} = \dfrac{m - n}{-(m - n)}$ = -1.

Substituting value of d in (i) we get,

⇒ a = n - m(-1) + (-1) = n + m - 1.

a_r = a + (r - 1)d

= n + m - 1 + (r - 1)(-1)

= n + m - 1 - r + 1

= n + m - r.

Hence, proved that r^th term is (m + n - r).