Problem 3 - Entrance Test

For an AP with first term a and common difference d, S_n = n/2 [2a + (n-1)d]. Given S_p = p/2 [2a + (p-1)d] = q ...(1) and S_q = q/2 [2a + (q-1)d] = p ...(2). Multiply (1) by q: pq/2 [2a+(p-1)d] = pq. Multiply (2) by p: pq/2 [2a+(q-1)d] = p^2. Equating: pq[2a+(p-1)d] = 2pq and pq[2a+(q-1)d] = 2p^2. From (1): 2a+(p-1)d = 2q/p. From (2): 2a+(q-1)d = 2p/q. Subtract: (p-1)d - (q-1)d = 2q/p - 2p/q → (p-q)d = 2(q^2 - p^2)/(pq) = -2(p^2-q^2)/(pq) = -2(p-q)(p+q)/(pq). Since p≠q, divide by (p-q): d = -2(p+q)/(pq). Now from (1): 2a = 2q/p - (p-1)d = 2q/p - (p-1)[-2(p+q)/(pq)] = 2q/p + 2(p-1)(p+q)/(pq). Simplify: 2a = 2q/p + 2(p^2-1)(p+q)/(p^2q)? Let me use a simpler approach. We need S_{p+q} = (p+q)/2 [2a + (p+q-1)d]. From the two equations, adding (1) and (2): p/2[2a+(p-1)d] + q/2[2a+(q-1)d] = p+q. Also, S_{p+q} = S_p + S_q + sum from term p+1 to p+q. The sum from term p+1 to p+q is q/2[2a_p + (q-1)d] where a_p = a+(p-1)d. This is messy. Better: From (1) and (2), we can find 2a+d(p+q-1). Note that 2a+(p-1)d = 2q/p and 2a+(q-1)d = 2p/q. Adding: 4a + (p+q-2)d = 2q/p + 2p/q. We want 2a+(p+q-1)d. Let me express S_{p+q} in terms of S_p and S_q. Using the standard result: If S_p = q and S_q = p, then S_{p+q} = -(p+q). This is a known result. Derivation: From (1): 2a+(p-1)d = 2q/p. From (2): 2a+(q-1)d = 2p/q. Multiply first by q and second by p: 2aq + (p-1)qd = 2q^2/p * q? Let me use elimination properly. From (1): 2a = 2q/p - (p-1)d. Substitute into (2): q/2[2(2q/p - (p-1)d) + (q-1)d] = p → q/2[4q/p - 2(p-1)d + (q-1)d] = p → q/2[4q/p - (2p-2-q+1)d] = p → q/2[4q/p - (2p-q-1)d] = p. This is too messy. Let me use the known shortcut: S_n = An^2 + Bn. Since S_n for AP is quadratic in n. Let S_n = An^2 + Bn. Then S_p = Ap^2 + Bp = q and S_q = Aq^2 + Bq = p. We need S_{p+q} = A(p+q)^2 + B(p+q). From the two equations: Ap^2 + Bp - q = 0 and Aq^2 + Bq - p = 0. Consider S_{p+q} + (p+q) = A(p+q)^2 + B(p+q) + (p+q) = (p+q)(A(p+q) + B + 1). Not helpful. Let me solve for A and B. From first: B = (q - Ap^2)/p. Substitute into second: Aq^2 + q(q - Ap^2)/p - p = 0 → Aq^2 + q^2/p - Aqp - p = 0 → A(q^2 - qp) = p - q^2/p → Aq(q-p) = (p^2 - q^2)/p = -(q^2-p^2)/p = -(q-p)(q+p)/p. So A = [-(q-p)(q+p)/p] / [q(q-p)] = -(q+p)/(pq). Since q-p = -(p-q), but we canceled. So A = -(p+q)/(pq). Then B = (q - Ap^2)/p = (q - [-(p+q)/(pq)]p^2)/p = (q + (p+q)p/q)/p = (q + p(p+q)/q)/p = (q^2 + p(p+q))/(pq) = (q^2 + p^2 + pq)/(pq). Now S_{p+q} = A(p+q)^2 + B(p+q) = (p+q)[A(p+q) + B] = (p+q)[-(p+q)^2/(pq) + (q^2+p^2+pq)/(pq)] = (p+q)/pq [-(p^2+2pq+q^2) + p^2+q^2+pq] = (p+q)/pq [-(p^2+2pq+q^2) + p^2+q^2+pq] = (p+q)/pq [-p^2-2pq-q^2+p^2+q^2+pq] = (p+q)/pq [-pq] = -(p+q). Therefore, S_{p+q} = -(p+q). Answer: -(p+q).