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

Answer

N/a

Explanation

Let the common difference be 'd'.
`"Given that "S_(p)=0" ...(1)"`
`rArr" "(p)/(2)[2a+(p-1)d]=0`
`rArr " "2a+(p-1)d=0`
`rArr " "d=(-2a)/(p-1)=(2a)/(1-p)" "...(2)`
Now, the sum next 'q' terms after first 'p' terms = sum of first (p+q) terms-sum of first p terms
`=(p+q)/(2)[2a+(p+q-1)d]-S_(p)`
`=(p+q)/(2)[2a+((p+q-1).2a)/(1-p)]-0`
[From eqs. (1) and (2)]
`=(p+q)/(2)[(2a(1-p)+(p+q-1).2a)/(1-p)]`
`=(p+q)/(2).2a[(1-p+p+q-1)/(1-p)]`
`=(a(p+q).q)/(1-p).`