If the pth,qth and rth terms of an A.P. be a,b and c respectively, show that : a(q-r)+b(r-p)+c(p-q)=0

Asked by Amuthavalli Thayyal Gopala Krishnan | 23rd Nov, 2013, 08:30: PM

Expert Answer:

Let A and d be ist term and common difference of the series AP
Then
a = A+(p-1).d.......(1)
b = A+(q-1).d.......(2)
c = A+(r-1).d........(3)
subtracting 2 from 1, 3 from 2 and 1st from 3rd we get
a-b = (p-q).d......(4)
b-c = (q-r).d........(5)
c-a = (r-p).d.......(6)
multiply 4,5,6 by c,a,b respectively we have
c.(a-b) = c.(p-q).d......(4)
a.(b-c) = a.(q-r).d........(5)
b.(c-a) = b.(r-p).d.......(6)
a(q-r).d+b(r-p).d+c(p-q).d = 0
(a(q-r)+b(r-p)+c(p-q)).d = 0
Now since d is common difference it should be non zero hence
a(q-r)+b(r-p)+c(p-q)= 0

Answered by  | 24th Nov, 2013, 12:07: AM

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