If a,b and c be the …

Given that, S_p = a, S_q = b and S_r = c
Let A be the first term and d be the common difference. Then,
S_p = {tex}\frac { p } { 2 }{/tex} [2 A + (p - 1)d] = a
{tex}\Rightarrow{/tex} 2 A + (p - 1)d = {tex}\frac { 2 a } { p }{/tex} ...(i)
S_q = {tex}\frac { q } { 2 }{/tex} [2 A + (q - 1)d] = b
{tex}\Rightarrow{/tex} 2 A + (q - 1)d = {tex}\frac { 2 b } { q }{/tex} ...(ii)
and S_r = {tex}\frac { r } { 2 }{/tex} [2 A + (r - 1)d] = c
{tex}\Rightarrow{/tex} 2 A + (r - 1)d = {tex}\frac { 2 c } { r }{/tex} ...(iii)
On multiplying Eq. (i) by (q - r), Eq. (ii) by (r - p) and Eq. (iii) by (p - q), we get
[2 A + (p - 1)d] (q - r) = {tex}\frac { 2 a } { p }{/tex} (q - r) ...(iv)
[2 A + (q - 1)d] (r - p) = {tex}\frac { 2 b } { q }{/tex} (r - p) ...(v)
and [2 A + (r - 1)d] (p - q) = {tex}\frac { 2 c } { r }{/tex} (p - q) ...(vi
On adding Eq. (iv), Eq. (v) and Eqs. (vi), we get
{tex}\frac { 2 a } { p } ( q - r ) + \frac { 2 b } { q } ( r - p ) + \frac { 2 c } { r } ( p - q ){/tex}
= [2 A + (p - 1)d] (q - r) + [2 A + (q - 1)d] (r - p) + [2 A + (r - 1)d] (p - q)
= 2 A (q - r + r - p + p - q) + d[(p - 1) (q - r) + (q - 1) (r - p) + (r - 1) (p - q)]
= 2 A (0) + d[(pq - pr - q + r + qr - qp - r + p + rp - rq - p + q]
= 0 + d (0) = 0
{tex}\therefore{/tex} {tex}\frac { a } { p } ( q - r ) + \frac { b } { q } ( r - p ) + \frac { c } { r } ( p - q ){/tex}= 0
Hence proved.