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

Let the first term be a and common difference be d.

Then,

T_p = a + (p − 1)d = q …(1)

T_q = a + (q − 1)d = p …(2)

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

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

⇒ (p − 1)d − (q − 1)d = q − p

⇒ pd - d - qd + d = q - p

⇒ pd - qd = q - p

⇒ (p − q)d = q − p

⇒ (p − q)d = −(p − q)

⇒ d = $\dfrac{−(p − q)}{(p − q)}$

⇒ d = -1.

Substituting d = −1 in equation (1), we get :

⇒ a + (p − 1)(−1) = q

⇒ a − p + 1 = q

⇒ a = p + q − 1.

Now, the (p + q)^th term :

T_{p + q} = a + (p + q − 1)d

= (p + q − 1) + (p + q − 1)(−1)

= (p + q − 1) − (p + q − 1)

= p - p + q - q - 1 + 1

= 0.

Hence, the (p + q)^th term of the A.P. is zero.