Arithmetic Progression
Question 2
Question 3 Important
Question 4
Question 5 Important
Question 6
Question 7 Important
Question 8
Question 9 Important
Question 10
Question 11 Important You are here
Question 12
Question 13
Question 14 Important
Question 15 Important
Question 16 Important
Question 17
Question 18 Important
Arithmetic Progression
Last updated at April 16, 2024 by Teachoo
Question11 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