Question

if a,b and c are the ,Qth, and Rth terms of A.P prove that a(q-r)+b(r-p)+ c(p-q)=0

Answer 1

Let's assume a is the pth term of the AP

As we know nth term An of AP=A+(n-1)d
A is the first term,
d is the common difference,
n is the number of terms.

Pth term Ap=a=A+(p-1)d----- eq1
qth term Aq=b=A+(q-1)d----- eq2
rth term Ar=c=A+(r-1)d-------- eq3

keeping the LHS in our view we need to multiply (q-r) , (r-p), (p-q) to eq1, 2 &3 respectively,

Let's multiply q-r to eq 1
a(q-r)=[A+(p-1)d](q-r)=A(q-r)+(p-1)(q-r)d --eq4

Let's multiply r-p to eq2
b(r-p)=[A+(q-1)d](r-p)=A(r-p)+(q-1)(r-p)d--eq5

Let's Multiply p-q to eq3
c(p-q)=[A+(r-1)d](p-q)=A(p-q)+(r-1)(p-q)d--eq6
Adding eq4, eq 5 and eq6
a(q-r)+b(r-p)+c(p-q)=A(q-r)+(p-1)(q-r)d+A(r-p)+(q-1)(r-p)d+A(p-q)+(r-1)(p-q)d
=A(q-r+r-p+p-q)+d[(p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)=A(0)+d(0) (On opening bracket)
=0=RHS( proved)