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The pth, qth and rth terms of an A.P. are a, b and c respectively. Show that a(q – r) + b(r - p) + c(p – q) = 0 |
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Answer» Let A be the first term and D the common difference of A.P. Tp = a = A + (p − 1)D = (A − D) + pD ... (1) Tq = b = A + (q − 1)D = (A − D) + qD ... (2) Tr = c = A + (r − 1)D = (A − D) + rD ...(3) Here we have got two unknowns A and D which are to be eliminated. We multiply (1), (2) and (3) by q−r, r−p and p−q respectively and add: a(q - r) = (A – D)(q - r) + Dp(q - r) b(r - p) = (A - D) (r - p) + Dq(r - p) c(p - q) = (A - D) (p - q) + Dr(p - q) a(q − r) + b(r − p) + c(p − q) = (A − D)[q − r + r − p + p − q] + D[p(q − r) + q(r − p) + r(p − q)] = (A – D) (0) + D [pq - pr + qr – pq + rp – rq] = 0 |
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