If p^th term of and A.P. is q and its q^th term is p, show that its r^th term is (p + q - r).

Let first term of A.P. be a and common term be d.

By formula,

⇒ a_n = a + (n - 1)d

Given,

⇒ p^th term = q

⇒ a + (p - 1)d = q ……….(1)

⇒ q^th term = p

⇒ a + (q - 1)d = p ……….(2)

Subtracting (2) from (1), we get :

⇒ a + (p - 1)d - [a + (q - 1)d] = q - p

⇒ a - a + (p - 1)d - (q - 1)d = (q - p)

⇒ d[p - 1 - (q - 1)] = (q - p)

⇒ d[p - 1 - q + 1] = (q - p)

⇒ d[p - q] = (q - p)

⇒ d[p - q] = -(p - q)

⇒ d = -1.

Substituting value of d in equation (1),

⇒ a + (p - 1)(-1) = q

⇒ a - p + 1 = q

⇒ a = p + q - 1.

r^th term = a_r

= a + (r - 1)d

= p + q - 1 + (r - 1)(-1)

⇒ p + q - 1 - r + 1

⇒ p + q - r.

Hence, proved that r^th term = p + q - r.