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Q.
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
Sequences and Series
Solution:
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}[2 a+(p-1) d] =\frac{q}{2}[2 a+(q-1) d] $
$\left\{\because S_n=\frac{n}{2}[2 a+(n-1) d]\right\}$
$\Rightarrow p[2 a+(p-1) d]=q[2 a+(q-1) d]$
$\Rightarrow 2 a p+p(p-1) d=2 a q+q(q-1) d$
$\Rightarrow 2 a p-2 a q=q(q-1) d-p(p-1) d$
$\Rightarrow 2 a(p-q)=d\left[q^2-q-p^2+p\right]$
$\Rightarrow 2 a(p-q)=d\left[(p-q)+\left(q^2-p^2\right)\right]$
$\Rightarrow 2 a(p-q)=d[(p-q)+(q-p)(q+p)]$
$\Rightarrow 2 a(p-q)=d(p-q)[1-(p+q)]$
$\Rightarrow 2 a=d[1-(p+q)]$
$\Rightarrow 2 a=-d[(p+q)-1]$
$\Rightarrow 2 a+d(p+q-1)=0 .....$(i)
Now, $ S_{p+q}=\frac{p+q}{2}[2 a+[p+q-1] d]$
$=\frac{p+q}{2} \times 0 $ [from Eq. (i)]
$=0$