Question

The first term of an AP is a and the sum of the first p terms is zero, show that the sum of its next q terms is $\frac{-a(p+q)q}{p-1}$

Answer 1

Let the common difference of an AP is d.

According to the question.

$S_p=0\Rightarrow \frac{P}{2}[2a+(p-1\left)d\right]=0[\because S_n=\frac{n}{2}\{2a+(n-1)d\left\}\right]\Rightarrow 2a+(p-1)d=0\therefore d=\frac{-2a}{p-1}\text{Now, sum of next q terms}=S_{p+q}-S_p=S_{p+q}-0=\frac{p+q}{2}[2a+(p+q-1\left)d\right]=\frac{p+q}{2}[2a+(p-1)d+qd]=\frac{p+q}{2}[2a+(p-1).\frac{-2a}{p-1}+\frac{q(-2a)}{p-1}]=\frac{p+q}{2}[2a+(-2a)-\frac{2aq}{p-1}]=\frac{p+q}{2}\left[\frac{-2aq}{p-1}\right]=\frac{-a(p+q)q}{(p-1)}$