If a, b and c are pth, qth and rth terms of an A.P., prove that a(q - r) + b(r − p) + c(p − q) = 0

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KnowledgeBoat · Accepted Answer

Let t and d be the first term and common difference of the A.P respectively. The nth term of an A.P is given by, an = t + (n - 1) d pth term = t + (p - 1)d = a ....(1) qth term = t + (q - 1)d = b ....(2) rth term = t + (r - 1)d = c ....(3) Subtracting (2) from (1), we obtain : ⇒ t + (p - 1)d - [t + (q - 1)d] = a - b ⇒ t + (p - 1)d - t - (q - 1)d = a - b ⇒ (p - 1 - q + 1)d = a - b ⇒ (p - q)d = a - b ⇒ d = ....(4) Subtracting (3) from (2), we obtain : ⇒ t + (q - 1)d - [t + (r - 1)d] = b - c ⇒ t + (q - 1)d - t - (r - 1) d = b - c ⇒ (q - 1 - r + 1)d = b - c ⇒ (q - r)d = b - c ⇒ d = ....(5) From (4) and (5), we get: ⇒ (a - b)(q - r ) = (b - c)( p - q) ⇒ aq - ar - bq + br = bp - bq - cp + cq ⇒ bp - cp + cq - aq + ar - br - bq + bq = 0 ⇒ (-aq + ar) + (bp - br) + (-cp + cq) = 0 ⇒ -a(q - r) - b(r - p) - c(p - q) = 0 ⇒ a(q - r) + b(r - p) + c(p - q) = 0 Hence, proved that a(q - r) + b(r − p) + c(p − q) = 0.

Mathematics

If a, b and c are p^th, q^th and r^th terms of an A.P., prove that
a(q - r) + b(r − p) + c(p − q) = 0

AP

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