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, $S_p = a$
$\Rightarrow \frac{p}{2}[2 A+(p-1) d]=a ....$(i)
$ S_q=b \Rightarrow \frac{q}{2}[2 A+(q-1) d]=b .....$(ii)
and $ S_r=c \Rightarrow \frac{r}{2}[2 A+(r-1) d]=c .....$(iii)
$ \left\{\because S_n=\frac{n}{2}[2 a+(n-1) d]\right\} $
Consider $=\frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q) $
$=\frac{1}{p} \times \frac{p}{2}[2 A+(p-1) d](q-r)+\frac{1}{q} \times \frac{q}{2}$
$ {[2 A+(q-1) d](r-p)+\frac{1}{r} \times \frac{r}{2}[2 A+(r-1) d](p-q)} $
[using Eqs. (i), (ii) and (iii)]
$=\frac{1}{2}[\{2 A+(p-1) d\}(q-r)+\{2 A+(q-1) d\}(r-p)$
$ +\{2 A+(r-1) d\}(p-q)]$
$=\frac{1}{2}[2 A(q-r)+(p-1) d(q-r)+2 A(r-p)$
$+(q-1) d(r-p)+2 A(p-q)+(r-1) d(p-q)]$
$=\frac{1}{2}[2 A(q-r+r-p+p-q)$
$+d[(p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)]$
$=\frac{1}{2}[2 A \times(0)+d(p q-p r-q+r$
$+q r-p q-r+p+r p-r q-p+q]$
$ = \frac{1}{2}(0 + d \times 0) = 0$