In an AP. Sp = q, Sq = p and Sr denotes the sum of first r terms. Then, Sp+q is equal to

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In the given problem, we are given S_p - q and S_q = p We need to find S_( p +q) Now, as we know, S_n = n/2 [ 2a + (n-1) d] So, S_p = p/2 [2a + (p-1) d] q = p/2 [2a + (p-1)d] 2q = 2ap + p (p - 1)d .................(1) Similarly, S_q = q/2 [2a + (q - 1) d ] p = q/2 [ 2a + (q - 1) d ] 2p = 2aq + q(q-1)d ......................(2) Subtracting (2) from (1), we get 2q - 2p = 2ap + [p(p - 1)d ] - 2aq - [q(q - 1 ) d ] 2q - 2p = 2a ( p - q) + [ p(p - 1 ) - q( q - 1 ) d -2(p-q) = 2a(p-q) + [(p2 - q2)-(p-q)] - 2 = 2a + ( p + q - 1 ) d ..................(3) Now, `S_(p + q) = ( p+q)/2 [2a + ( p+q - 1 ) d ] S_(p+q) = ((p+q))/2 (-2) ....(Using 3 ) S_(p+q) = - (p+ q) Thus, S_(p+q) = - (p+ q) `

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