# Misc 15

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

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

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

Misc 22

Misc 23

Misc 24

Misc 25 Important

Misc 26

Misc 27

Misc 28 Important

Misc 29

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Misc 31

Misc 32 Important

Chapter 9 Class 11 Sequences and Series

Serial order wise

About the Author

CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .