Question

Sum of the first $p,q$ and $r$ terms of an A.P. are $a,b$ and $c$, respectively. Then,

• A. $\frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)=0$
• B. $\frac{a}{p}(q+r)+\frac{b}{q}(r+p)+\frac{c}{r}(p+q)=0$
• C. $\frac{a}{p}(q-r)-\frac{b}{q}(r-p)-\frac{c}{r}(p-q)=0$
• D. $\frac{a}{p}(q+r)-\frac{b}{q}(r+p)-\frac{c}{r}(p+q)=0$

Answer 1

Answer: (A)

Let the first term is $A$ and common difference is $d$.
Given, $S_p=a$
$\Rightarrow \frac{p}{2}[2A+(p-1)d]=a....$(i)
$S_q=b\Rightarrow \frac{q}{2}[2A+(q-1)d]=b.....$(ii)
and $S_r=c\Rightarrow \frac{r}{2}[2A+(r-1)d]=c.....$(iii)
$\left\{\because S_n=\frac{n}{2}[2a+(n-1)d]\right\}$
Consider $=\frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)$
$=\frac{1}{p}\times \frac{p}{2}[2A+(p-1)d](q-r)+\frac{1}{q}\times \frac{q}{2}$
$[2A+(q-1)d](r-p)+\frac{1}{r}\times \frac{r}{2}[2A+(r-1)d](p-q)$
[using Eqs. (i), (ii) and (iii)]
$=\frac{1}{2}[\{2A+(p-1)d\}(q-r)+\{2A+(q-1)d\}(r-p)$
$+\{2A+(r-1)d\}(p-q)]$
$=\frac{1}{2}[2A(q-r)+(p-1)d(q-r)+2A(r-p)$
$+(q-1)d(r-p)+2A(p-q)+(r-1)d(p-q)]$
$=\frac{1}{2}[2A(q-r+r-p+p-q)$
$+d[(p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)]$
$=\frac{1}{2}[2A\times (0)+d(pq-pr-q+r$
$+qr-pq-r+p+rp-rq-p+q]$
$=\frac{1}{2}(0+d\times 0)=0$