Question

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

Answer 1

$\mathfrak\green{Answer \ ♡ }$

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)

Tr = 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