Question

In an A.P. of which $a$ is the first term, if the sum of the first $p$ terms is zero, then the sum of next $q$ terms is

• A. $-\frac{a(p+q)p}{q+1}$
• B. $\frac{a(p+q)p}{p+1}$
• C. $-\frac{a(p+q)q}{p-1}$
• D. None of the above

Answer 1

The correct option is C $-\frac{a(p+q)q}{p-1}$

Let $d$ be the common difference.
Given, $S_p=0$
$\therefore \frac{p}{2}[2a+(p-1\left)d\right]=0\Rightarrow d=-\frac{2a}{p-1}$

Sum of next $q$ terms is sum of an A.P. whose first term will be $T_{p+1}=a+pd$
$\therefore S=\frac{q}{2}\left[2\right(a+pd)+(q-1\left)d\right]=\frac{q}{2}[2a+(p-1)d+(p+q\left)d\right]=\frac{q}{2}[0-(p+q\left)\frac{2a}{p-1}\right]=-\frac{a(p+q)q}{p-1}$