The pth, qth and rth term of an AP are a, b and c respectively. Show that a(q-r)+b(r-p)+c(p-q)=0
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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.
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