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

Given: a_p = q

a_q = p

To show: a_{(p + q)} = 0

Consider a_p = q

⇒ a + (p - 1)d = q …………………(1)

Consider a_q = p

⇒ a + (q - 1)d = p ………………….(2)

Now, subtracting equation (2) from equation (1), we get

(p - q)d = (q - p)

⇒ d = - 1

∴ From equation (1), we get,

a - p + 1 = q

⇒ p + q = a + 1 ……………………….(3)

Consider a_{(p + q)} = a + (p + q - 1)d

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

= a + (a + 1 - 1)(-1)

(putting the value of p + q from equation 3)

= a + (-a)

= 0

∴ a_{(p + q)} = 0

Hence, proved.