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# Sum of first $p,q\:\:and\:\:r\:terms$ of an $A.P.$ are $a,b\:\:and\:\:c$ respectively. Prove that $\large\frac{a}{p}$$(q-r)+\large\frac{b}{q}$$(r-p)+\large\frac{c}{r}$$(p-q)=0 Can you answer this question? ## 1 Answer 0 votes Toolbox: • S_n=\large\frac{n}{2}$$[2A+(n-1)d]$ where $A$ is first term and $d$ is common difference.
Given: Sum of $1^{st}\:p\:terms=S_p=a,$
Sum of $1^{st}\:q\:terms=S_q=b\:\:and\:\:$Sum of $1^{st}\:r\:terms=S_r=c,$
$\Rightarrow\:\large\frac{p}{2}$$[2A+(p-1)d]=a........(i) \large\frac{q}{2}$$[2A+(q-1)d]=b........(ii)$ and
$\large\frac{r}{2}$$[2A+(r-1)d]=c........(iii) where A is 1^{st} term and d is common difference. Substituting the value of a in \large\frac{a}{p}$$(q-r)$
$\Rightarrow\:\large\frac{a}{p}$$(q-r)=\large\frac{\large\frac{p}{2}[2A+(p-1)d]}{p}$$(q-r)$
$=\bigg[A+\large\frac{p-1}{2}$$d\bigg](r-p) Similarly by substituting the value of b in \large\frac{b}{q}$$(r-p)$
$\Rightarrow\:\large\frac{b}{q}$$(r-p)=\large\frac{\large\frac{q}{2}[2A+(q-1)d]}{q}$$(r-p)$
$=\bigg[A+\large\frac{q-1}{2}$$d\bigg](r-p) and by substituting the value of c in \large\frac{c}{r}$$(p-q)$
$\Rightarrow\:\large\frac{c}{r}$$(p-q)=\large\frac{\large\frac{r}{2}[2A+(r-1)d]}{r}$$(p-q)$
$=\bigg[A+\large\frac{r-1}{2}$$d\bigg](p-q) \therefore\:\large\frac{a}{p}$$(q-r)+\large\frac{b}{q}$$(r-p)+\large\frac{c}{r}$$(p-q)=$
$=\bigg[A+\large\frac{p-1}{2}$$d\bigg](q-r)+\bigg[A+\large\frac{q-1}{2}$$d\bigg](r-p)+\bigg[A+\large\frac{r-1}{2}$$d\bigg](p-q) =A[q-r+r-p+p-q]+\large\frac{d}{2}$$\big[pq-q-pr+r+qr-r-pq+p+pr-p-qr+q\big]$