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

Given: [ a_p = q \implies a + (p-1)d = q ] [ a_q = p \implies a + (q-1)d = p ]

Subtract: [ (p-1 - (q-1))d = q - p ] [ (p - q)d = q - p = -(p - q) ] [ d = -1 \quad (\text{assuming } p \ne q) ]

Then [ a + (p-1)(-1) = q \implies a - p + 1 = q \implies a = p + q - 1 ]

Now ( n^{th} ) term: [ a_n = a + (n-1)d = (p + q - 1) + (n-1)(-1) = p + q - 1 - n + 1 = p + q - n ]

Proved.