Question

If the pth, qth and rth terms of an A.P. be a, b and c respectively. Then prove that a(q–r) + b(r–p) + c(p–q) = 0.

Answer 1

Hint: We will use the nth term of an arithmetic progression (A.P) with first term ‘a’ and common difference ‘d’ is $T_n=a+(n-1)d$.

Complete step-by-step answer:
Let A be the first term D be the common difference of A.P.
The pth term is $T_p=a=A+(p-1)D=(A-D)+pD\to (1)$
The qth term is $T_q=b=A+(q-1)D=(A-D)+qD\to (2)$
The rth term is$T_r=c=A+(r-1)D=(A-D)+rD\to (3)$
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 them together.
$$\Rightarrow a(q-r)+b(r-p)+c(p-q)$$
Substituting a, b and c values from equations (1), (2) and (3)
$$\Rightarrow \left((A-D)+pD\right)(q-r)+\left((A-D)+qD\right)(r-p)+\left((A-D)+rD\right)(p-q)$$
On simplification,
$$\Rightarrow \left(A-D\right)\left[q-r+r-p+p-q\right]+D\left[p(q-r)+q(r-p)+r(p-q)\right]$$
$$\begin{array}{rl}& \Rightarrow \left(A-D\right)\cdot (0)+D\left[pq-pr+qr-pq+pr-qr\right]\\ & \Rightarrow 0+D\cdot 0\end{array}$$
= 0
$$\therefore a(q-r)+b(r-p)+c(p-q)=0$$
Hence proved.

Note: We wrote pth, qth and rth terms of A.P. using nth term formula. Then we did algebraic calculation to prove the given condition.