Let in an AP, first term is a and common difference is d

Now S_p = q

=> (p/2) *{2a + (p-1)d} = q

=> p*{2a + (p-1)d} = 2q

=> 2ap + p² d - pd = 2q ................1

Again S_q = p

=> (q/2) *{2a + (q-1)d] = p

=> q*{2a + (q-1)d} = 2p

=> 2aq + q² d - qd = 2p .........2

subtract equation1 and 2, we get

      2ap + p² d - pd - (2aq + q² d - qd) = 2q - 2p

=> 2ap + p² d - pd - 2aq - q² d + qd = 2q - 2p

=> 2a(p-q) + (p² - q² )d - d (p-q) = -2(p-q)

=> 2a(p-q) + (p - q)*(p+q)*d - d (p-q) = -2(p-q)

=> 2a + (p+q)d - d = -2                                           (divide by (p-q) on both side)

=> 2a + (p+q -1)d = -2

Now multiply (p+q)/2 on both side, we get

      (p+q)/2*{2a + (p+q -1)d} = -2*(p+q)/2

=> S_p+q = -(p+q)