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Last updated at May 29, 2018 by Teachoo

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Misc 15 The pth, qth and rth terms of an A.P. are a, b, c respectively. Show that (q r) a+(r p)b+ (p q) c = 0 Here we have small a in the equation, so we use capital A for first term We know that, An = A + (n 1) D where An is the nth term of A.P. n is the number of terms A is the first term, D is the common difference It is given that pth term of an AP is a i.e. Ap = a Putting n = p A + (p 1) D = a a = A + (p 1)D qth term of an AP is b i.e. Aq = b Putting n = q A + (q 1)D = b b = A + (q 1)D rth term of an AP is c i.e. Ar = c Putting n = r A + (r 1)D = c c = A + (r 1)D Now we need to show that (q r)a + (r p)b + (p q)c = 0 We have a = A + (p 1)D Multiplying by (q r) a (q r) = [A + (p 1)D] (q r) a (q r) = [A (q r) + (q r) (p 1)D] And b = [A + (q 1)D] Multiplying by (r p) b (r p) = [A + (q 1)D] (r p) b (r p) = [A (r p) + (q r) (q 1)D] Similarly, c = A + (r 1)D multiply this by (p q) c (p q) = [A + (r 1)D] (p q) c (p q) = [A (p q) + (p q) (r 1)D] We have to prove a (q r) + b (r p) + c (p q) = 0 Taking L.H.S a (q r) + b (r p) + c (p q) From (1), (2) & (3) = "[A (q r) + (q r) (p 1)D] + [A (r p) + (r p) (q 1)D]" "+ [A (p q) + (p q) (r 1)D]" = "A (q r) + A (r p) + A (p q) + (q r) (p 1)D" "+ (r p) (q 1)D + (p q) (r 1)D" = A[q r + r p + p q] + D[(q r) (p 1) + (r p) (q 1) + (p q) (r 1)] = A[p p + r r + q q] + D[q(p 1) r(p 1) + r(q 1) p(q 1) + p(r 1) q(r 1)] = A[0 + 0 + 0] + D[qp q rp + r + rq r pq + p + pr p qr + q] = 0 + D[qp pq + rq qr rp + pr + q q + r r + p p ] = 0 + D[0] = 0 = R.H.S Hence proved

Miscellaneous

Misc 1

Misc 2

Misc 3 Important

Misc 4

Misc 5

Misc 6

Misc 7 Important

Misc 8

Misc 9

Misc 10

Misc 11

Misc 12

Misc 13

Misc 14

Misc 15 You are here

Misc 16 Important

Misc 17

Misc 18

Misc 19 Important

Misc 20

Misc 21 Not in Syllabus - CBSE Exams 2021

Misc 22

Misc 23 Not in Syllabus - CBSE Exams 2021

Misc 24 Not in Syllabus - CBSE Exams 2021

Misc 25 Important Not in Syllabus - CBSE Exams 2021

Misc 26

Misc 27

Misc 28 Important

Misc 29

Misc 30

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Misc 32 Important

Chapter 9 Class 11 Sequences and Series

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.