Let the first term be A & common difference be D. We have
Sp =

p

2

[2A + (p - 1) D] = a ⇒ A + (p - 1)

D

2

=

a

p

, Sq =

q

2

[2A + (q - 1) D] = b
⇒ A + (q - 1)

D

2

=

b

q

, Sr =

r

2

[2A + (r - 1) D] = c ⇒ A + (r - 1)

D

2

=

c

r

Now,

a

p

(q−r)+b/q$$ (r - p) +

c

r

(p - q)
= (A+(p−1)

D

2

)(q−r) + (A+(q−1)

D

2

)(r−p) + (A+r−1)

D

2

)(p−q)
= A [(q - r) + (r - p) + (p - q)] +

D

2

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

D

2

[pq - rp - q + r + qr - pq - r + p + rp - qr - p + q]
= A (0) +

D

2

(0) = 0 + 0 = 0
Hence

q

p

(q - r) +

b

q

(r - p) +

c

r

(p - q) = 0