If p^{th} , q^{th} and r^{th} terms of an A.P. are a, b, c respectively, then show that

(i) a (q – r) + b(r – p) + c(p – q) = 0

(ii) (a – b) r + (b –¬ c) p + (c – a) q = 0

#### Solution

Let A be the first term and D be the common difference of the given A.P. Then,

a = p^{th} term ⇒ a = A + (p – 1) D ….(i)

b = q^{th} term ⇒ b = A + (q – 1) D ….(ii)

c = r^{th} term ⇒ c = A+ (r – 1) D ….(iii)

(i) We have,

a(q – r) + b (r – p) + c (p – q)

= {A + (p – 1) D} (q – r) + {A + (q – 1)} (r – p) + {A + (r – 1) D} (p – q)

[Using equations (i), (ii) and (iii)]

= A {(q – r) + (r – p) + (p – q)} + D {(p – 1) (q – r) + (q – 1) (r – p) + (r – 1) (p – q)}

= A {(q – r) + (r – p) + (p – q)} + D{(p – 1) (q – r) + (q – 1) (r – p) + (r – 1) (p – q)}

= A . 0 + D {p (q – r) + q (r – p) + r (p – q) – (q – r) – (r – p) – (p– q)}

= A . 0 + D . 0 = 0

(ii) On subtracting equation (ii) from equation (i), equation (iii) from equation (ii) and equation (i) from equation (iii), we get

a – b = (p – q) D, (b – c) = (q – r) D and c – a = (r – p) D

∴ (a – b) r + (b – c) p + (c – a) q

= (p – q) Dr + (q – r) Dp + (r – p) Dq

= D {(p – q) r + (q – r) p + (r – p) q}

= D × 0 = 0