The first term of an A.P. is `a` and the sum of first `p` terms is zero, show tht the sum of its next `q` terms is `(a(p+q)q)/(p-1)dot`

Explanation

let the common dfference of an AP is d.
According to the question
`s_(p) =0`
`implies (p)/(2)[2a+(p-1)d]=0 " "[:' s_(n) =(n)/(2){2a+(n-1)d}]`
` implies 2a+(p-1)d=0`
`d=(-2a)/(p-1)`
Now , sum of next q terms `=s_(p+q)-S_(p)=S_(p-q)-0`
`=(p+q)/(2)[2a+(p+q-1)d]`
`=(p+q)/(2)[2a+(p-1)d+qd]`
`=(p+q)/(2)[2a+(p-1).(-2a)/(p-1)+(q(-2))/(p-1)]`
`=(p+q)/(2)[2a+(-2a)-(2aq)/(p-1)]`
`=(P+q)/(2)[(-2aq)/(p-1)]`
`=(-a(P+q)q)/((p-1))`