Question

If the sum of first $p$ terms of an A.P. is equal to the sum of first $q$ terms, then sum of first $(p+q)$ terms i.e., $S_{p+q}$ is

• A. 1
• B. 2
• C. 0
• D. 3

Answer 1

Answer: (C)

Let $a$ and $d$ be the first term and common difference of an A.P.
$\because$ Sum of the first $p$ terms $=$ Sum of first $q$ terms
$\therefore S_p=S_q$
$\Rightarrow \frac{p}{2}[2a+(p-1)d]=\frac{q}{2}[2a+(q-1)d]$
$\left\{\because S_n=\frac{n}{2}[2a+(n-1)d]\right\}$
$\Rightarrow p[2a+(p-1)d]=q[2a+(q-1)d]$
$\Rightarrow 2ap+p(p-1)d=2aq+q(q-1)d$
$\Rightarrow 2ap-2aq=q(q-1)d-p(p-1)d$
$\Rightarrow 2a(p-q)=d\left[q^2-q-p^2+p\right]$
$\Rightarrow 2a(p-q)=d\left[(p-q)+\left(q^2-p^2\right)\right]$
$\Rightarrow 2a(p-q)=d[(p-q)+(q-p)(q+p)]$
$\Rightarrow 2a(p-q)=d(p-q)[1-(p+q)]$
$\Rightarrow 2a=d[1-(p+q)]$
$\Rightarrow 2a=-d[(p+q)-1]$
$\Rightarrow 2a+d(p+q-1)=\mathrm{0.....}$(i)
Now, $S_{p+q}=\frac{p+q}{2}[2a+[p+q-1]d]$
$=\frac{p+q}{2}\times 0$ [from Eq. (i)]
$=0$