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.

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

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,
$\Rightarrow \left( {A - D} \right)\left[ {q - r + r - p + p - q} \right] + D\left[ {p(q - r) + q(r - p) + r(p - q)} \right]$
\eqalign{ & \Rightarrow \left( {A - D} \right) \cdot (0) + D\left[ {pq - pr + qr - pq + pr - qr} \right] \cr & \Rightarrow 0 + D \cdot 0 \cr}
= 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.