Sum of the first p, q and r terms of an A.P are a, b and c, respectively.Prove that `a/p(q-r)+b/q(r-p)+c/r(p-q)=0`

Explanation

Let the first term and common difference of A.P. be x and y respectively.
`:. S_(p)=a`
`rArr (p)/(2)[2x+(p-1)y]=a`
`rArr (a)/(p)=x+(p-1)(y)/(2)`
`rArr (a)/(p)(q-r)=x(q-r)+(p-1)(q-r)(y)/(2) " " ...(1)`
Similarly, `S_(q)=b`
`rArr (q)/(2)[2x+(q-1)y]=b`
`rArr (b)/(q)=x+(q-1)(y)/(2)`
`rArr (b)/(q)(r-p)=x(r-p)+(q-1)(r-p)(y)/(2) " " ...(2)`
and `S_(r)=c`
`rArr (r)/(2)[2x+(r-1)y]=c`
`rArr (c)/(r)=x+(r-1)(y)/(2)`
`rArr (c)/(r)(p-q) = x(p-q)+(r-1)(p-q)(y)/(2) " " ...(3)`
Adding eqs. (1), (2) and (3), we get
`(a)/(p)(q-1)+(b)/(q)(r-p)+(c)/(r)(p-q)=x{(q-r)+(r-p)+(p-q)}+(y)/(2){(p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)}`
`= 0 +(y)/(2){pq-pr-q+r+qr-pq-r+p+rp-rq-p+q}`
`=0 " "` Hence Proved.