Question

If the sum of an AP is the same for $p$ terms as for the $q$ terms , Find the sum for $(p+q)$ terms

• A. $2$
• B. $0$
• C. $4$
• D. None of these

Answer 1

Answer: (B)

$0$

Since $S_p=S_q$

$\frac{p}{2}[2a+(p-1\left)d\right]=\frac{q}{2}[2a+(q-1\left)d\right]2ap+(p-1)pd=2aq+q(q-1)d2a(p-q)=\left[q\right(q-1)-p(p-1\left)\right]d2a(p-q)=[q^2-p^2+(p-q\left)\right]d2a(p-q)=\left[\right(q-p\left)\right(q+p)+(p-q\left)\right]d2a=[-(q+p)+1]d-2a=(q+p-1)d\therefore S_{p+q}=\frac{(p+q)}{2}[2a+(p+q-1\left)d\right]S_{p+q}=\frac{(p+q)}{2}(2a-2a)=0$