Question

If the sum of first $p$ terms of an A.P. is equal to the sum of first $q$ terms, then find the sum of the first $(p+q)$ terms.

• A. $3$
• B. $2$
• C. $7$
• D. $0$

Answer 1

Answer: (D)

$0$

Let $a$ and $d$ be the first term and common difference of the A.P. respectively.

Here,

$S_p=\frac{p}{2}[2a+(p-1\left)d\right]$

$S_q=\frac{q}{2}[2a+(q-1\left)d\right]$

According to the question,

$\frac{p}{2}[2a+(p-1\left)d\right]=\frac{q}{2}[2a+(q-1\left)d\right]$

$p[2a+(p-1\left)d\right]=q[2a+(q-1\left)d\right]$

$2ap+pd(p-1)=2aq+qd(q-1)$

$2a(p-q)+d\left[p\right(p-1)-q(q-1\left)\right]=0$

$2a(p-q)+d[p^2-p-q^2+q]=0$

$2a(p-q)+d\left[\right(p-q\left)\right(p+q-1\left)\right]=0$

$2a+d(p+q-1)=0$

$d=-\frac{2a}{p+q-1}$

Therefore, $S_{p+q}=\frac{p+q}{2}[2a+(p+q-1\left)d\right]$

$S_{p+q}=\frac{p+q}{2}[2a+(p+q-1\left)\right(-\frac{2a}{p+q-1}\left)\right]$

$=\frac{p+q}{2}[2a-2a]=0$