The sum of the first p, q, r terms of an A.P. are a, b, c respectively. Show that `\frac { a }{ p } (q – r) + \frac { b }{ q } (r – p) + \frac { c }{ r } (p – q) = 0`

#### Solution

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

a = Sum of p terms ⇒ a = `\frac { p }{ 2 } [2A + (q – 1) D]`

`⇒ \frac { 2a }{ p } = [2A + (p – 1) D] ….(i)`

b = Sum of q terms

`⇒ b = \frac { q }{ 2 } [2A + (q – 1) D]`

`⇒ \frac { 2b }{ q } = [2A + (q – 1) D] ….(ii)`

and, c = Sum of r terms

`⇒ c = \frac { r }{ 2 } [2A + (r – 1) D]`

`⇒ \frac { 2c }{ r } = [2A + (r – 1) D] ….(iii)`

Multiplying equations (i), (ii) and (iii) by (q – r), (r – p) and (p –q) respectively and adding, we get

`\frac { 2a }{ p } (q – r) + \frac { 2b }{ q } (r – p) + \frac { 2c }{ r} (p – q)`

`= [2A + (p – 1) D] (q – r) + [2A + (q – 1) D] (r – p) + [(2A + (r – 1)D] (p – q)`

= 2A (q – r + r – p + p – q) + D [(p – 1) (q – r) + (q – 1)(r – p) + (r– 1) (p – q)]

= 2A × 0 + D × 0 = 0