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

then sum of n terms in A.P is S_n = ()[ 2a + (n – 1) d]

Here, s_p= q

S_q =p

S_p = () [2a + (p – 1) d]

q = [2a + (p – 1) d]

= [2a + (p – 1) d] --------(1)

S_q = p = ( [2a + (q – 1) d]

= [2a + (q – 1) d] …………..(2)

Subtract (1) from (2) we get

(q – p) d = / pq -----------(3)

d = -2(q + p) / pq -----------(3)

Sum of first (p + q) terms

[from (1) and (3)]