Question

If a, b, c are the pth, qth, and rth terms of an AP respectively, show that a(q-r) +b(r-p) +c(p-q) =0

Answer 1

Let a=first term of an A.P.
and
Let d=common difference of A.P.

Now,
a=A+(p-1).d -----(1)
b=A+(q-1).d------(2)
c=A+(r-1).d ------(3)

Subtracting 2nd from 1st equation , 3rd from 2nd equation and 1st from 3rd equation
and we get,

a-b=(p-q).d -----(4)
b-c=(q-r).d --------(5)
c-a=(r-p).d--------(6)

Multiply equation 4,5, and 6 by c,a and b
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)}=0

Now, since d is common difference it should be non zero

So, a(q-r) + b (r-p) + c(p-q)=0

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