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`

Options

A. `(-a(p+q)p)/(q+1)`
B. `(a(q+q)p)/(P+1)`
C. `(-a(p+q)q)/(p-1)`
D. none of these

Answer

C

Explanation

Given, `S_(p)=0`. Therefore,
`p/2[2a+(p-1)d]=0` or `d=(-2a)/(p-1)` (1)
Sum of next q terms is sum of an A.P. whose first term will be
`T_(p+1)=a+pd`
`thereforeS=q/2[2(a+pd)+(q-1)d]`
`=q/2[2a+(p-1)d+(p+q)d]`
`=q/2[0-(p+q)(2a)/(p-1)]`
`=-a((p+q)q)/(p-1)` [Using (1)]