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Ex 9.2, 11 - Sum of first p, q, r terms of AP are a, b, c - Arithmetic Progression (AP): Calculation based/Proofs

Ex 9.2, 11 - Chapter 9 Class 11 Sequences and Series - Part 2
Ex 9.2, 11 - Chapter 9 Class 11 Sequences and Series - Part 3 Ex 9.2, 11 - Chapter 9 Class 11 Sequences and Series - Part 4 Ex 9.2, 11 - Chapter 9 Class 11 Sequences and Series - Part 5 Ex 9.2, 11 - Chapter 9 Class 11 Sequences and Series - Part 6 Ex 9.2, 11 - Chapter 9 Class 11 Sequences and Series - Part 7

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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 = /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 /2 [2A + (p 1)D] = a a = /2 [2A + (p 1)D] Similarly, Sum of first q terms = b Sq = b /2 [2A + (q 1)D] = b b = /2 [2A + (q 1)D] Similarly, Sum of first r terms = c Sr = c /2 [2A + (r 1)D] = c c = /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] / = 1/2 [2A + (p 1)D] / = 1/2 (2A) + 1/2 (p 1)D / = A + (( 1))/2D Multiplying both sides by q r / (q r) = ("A + " (( 1))/2 "D" )(q r) Similarly, b = q/2 [2A + (q 1)D] / = 1/2 [2A + (q 1)D] / = 1/2 (2A) + 1/2 (q 1)D / = A + (( 1))/2D Multiplying both sides by r p / (r p) = ("A + " (( 1))/2 "D" )(r p) Similarly, c = r/2 [2A + (r 1)D] / = 1/2 [2A + (r 1)D] / = 1/2 (2A) + 1/2 (r 1)D / = A + (( 1))/2D Multiplying both sides by p q " " / "(p q) =" ("A + " (( 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 + " (( 1))/2 "D" )"(q r) +" ("A + " (( 1))/2 "D" )"(r p) +" ("A + " (( 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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.