Last updated at Dec. 8, 2016 by Teachoo

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Ex 9.2,11 Sum of first p,q,r terms of an A.P are a,b,c resp. "Prove that" a/p " (q − r) + " b/q " (r − p)+ " c/r " (p − q) = 0" Here we have small ‘a’ in the equation, so we use capital ‘A’ for first term We know that, Sn = 𝑛/2 [2A + (n – 1)D] where Sn is the sum of n terms of A.P. n is the number of terms A is the first term, D is the common difference Given, Sum of first p terms = a Sp = a 𝑝/2 [2A + (p – 1)D] = a a = 𝑝/2 [2A + (p – 1)D] Similarly, Sum of first q terms = b Sq = b 𝑞/2 [2A + (q – 1)D] = b b = 𝑞/2 [2A + (q – 1)D] Similarly, Sum of first r terms = c Sr = c 𝑟/2 [2A + (r – 1)D] = c c = 𝑟/2 [2A + (r – 1)D] We need to "prove that" a/p " (q − r) + " b/q " (r − p)+ " c/r " (p − q) = 0" We have a = p/2 [2A + (p – 1)D] 𝑎/𝑝 = 1/2 [2A + (p – 1)D] 𝑎/𝑝 = 1/2 (2A) + 1/2 (p – 1)D 𝑎/𝑝 = A + ((𝑝 − 1))/2D Multiplying both sides by q – r 𝑎/𝑝(q – r) = ("A + " ((𝑝 − 1))/2 "D" )(q – r) Similarly, b = q/2 [2A + (q – 1)D] 𝑏/𝑞 = 1/2 [2A + (q – 1)D] 𝑏/𝑞 = 1/2 (2A) + 1/2 (q – 1)D 𝑏/𝑞 = A + ((𝑞 − 1))/2D Multiplying both sides by r – p 𝑏/𝑞(r – p) = ("A + " ((𝑞 − 1))/2 "D" )(r – p) Similarly, c = r/2 [2A + (r – 1)D] 𝑐/𝑟 = 1/2 [2A + (r – 1)D] 𝑐/𝑟 = 1/2 (2A) + 1/2 (r – 1)D 𝑐/𝑟 = A + ((𝑟 − 1))/2D Multiplying both sides by p – q " " 𝑐/𝑟 "(p – q) =" ("A + " ((𝑟 − 1))/2 "D" )(p – q) We need to "prove that" a/p " (q − r) + " b/q " (r − p)+ " c/r " (p − q) = 0" Taking L.H.S a/p " (q − r) + " b/q " (r − p)+ " c/r " (p − q)" Putting values from (1), (2) & (3) = ("A + " ((𝑝 − 1))/2 "D" )"(q – r) +" ("A + " ((𝑞 − 1))/2 "D" )"(r – p) +" ("A + " ((𝑟 − 1))/2 "D" )"(p – q)" = "[A(q – r) + " 1/2 "(p – 1)(q – r)D]" +"[A(r – p) +" 1/2 "(q – 1)(r – p)D]" "+ [A(p – q) + " 1/2 "(r – 1)(p − q)D]" = "[A(q – r) + A(r – p) + A(p – q)]" █(" " @"+ " 1/2 " D[(p – 1)(q – r) + (q – 1)(r – p) + (r – 1)(p – q)] " ) = "A(q – r + r – p + p – q)" "+ " 1/2 " D[pq – pr− q + r + qr – pq –r + p +rp – rq – p + q]" = █("A(q – q – r + r + p – p) " ) "+ " 1/2 " D[pq – qp + pr – rp – q + p + r – r + p – q + qr – rq]" = █("A(0) " ) "+ " 1/2 " D[0]" = 0 + 0 = 0 = R.H.S Thus , L.H.S = R.H.S Hence proved

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.