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

Explanation

Let A be the first term and D the common difference of A.P.

`T _p` `=a=A+(p−1)D=(A−D)+pD .. (1)`

`T _q``=b=A+(q−1)D=(A−D)+qD ..(2)`

`T _r` `=c=A+(r−1)D=(A−D)+rD ..(3)`

Here we have got two unknowns A and D which are to be eliminated.

We multiply (1),(2) and (3) by `q−r,r−p` and `p−q `respectively and add:

`a(q−r)+b(r−p)+c(p−q)`

`=(A−D[q−r+r−p+p−q]+D[p(q−r)+q(r−p)+r(p−q)]=0.`