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`

Answer

N/a

Explanation

Let first term be x and common difference be y of A.P.
According to the problem, `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)cdot(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)"`
`"Similarly, "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 equations (1), (2) and (3)
`(a)/(p)(q-r)+(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-qp-r+p+rp-rp-p+q)`
`=0`.