Question

please solve this .......​

Answer 1

Sol: Let a be the first term and d is common difference of the A.P then sum of n terms in A.P is Sn = (n/2)[ 2a + (n - 1) d] Given that Sp = q and Sq = p. Sp = (p/2)[ 2a + (p - 1) d] = q ⇒ [ 2a + (p - 1) d] = 2q / p --------(1) Sq = (q/2)[ 2a + (q - 1) d] = p ⇒ [ 2a + (q - 1) d] = 2p / q --------(2) Subtract (1) from (2) we get (q - p)d = (2p / q) – (2q / p) (q - p)d = (2p2– 2q2) / pq d = -2(q +p) / pq -----------(3) Sum of first ( p + q ) terms Sp +q = (p+ q) / 2 [ 2a + ( p + q -1) d] Sp +q = (p+ q) / 2 [ 2a + ( p -1)d + qd] Sp +q = (p +q) / 2 [ (2q /p) + q(-2(q +p) / pq )] [ from (1) and (3)] Sp +q = (p + q) / 2 [ (2q -2q-2p) / p )] Sp +q = (p + q) / 2 [-2p) / p )] Sp +q = - (p + q).