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# Sum of the first p, q and r terms of an A.P. are a, b and c, respectively. Prove that (Q11)

11. Sum of the first p, q and r terms of an A.P. are a, b and c, respectively. Prove that

$\frac{a}{p} ( q-r ) + \frac{b}{q}( r-p ) + \frac{c}{r} ( p-q ) = 0$

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To prove : $\frac{a}{p} ( q-r ) + \frac{b}{q}( r-p ) + \frac{c}{r} ( p-q ) = 0$

Let $a_1$ and d be the first term and the common difference of AP, respectively.

According to the given information, we have

$S_p=\frac{p}{2}[2a_1+(p-1)d]=a$

$\Rightarrow [2a_1+(p-1)d]=\frac{2a}{p}............(1)$

$S_q=\frac{q}{2}[2a_1+(q-1)d]=b$

$\Rightarrow [2a_1+(q-1)d]=\frac{2b}{q}............(2)$

$S_r=\frac{r}{2}[2a_1+(r-1)d]=c$

$\Rightarrow [2a_1+(r-1)d]=\frac{2c}{r}............(3)$

Subtracting equation (2) from (1), we have

$\Rightarrow (p-1)d-(q-1)d=\frac{2a}{p}-\frac{2b}{q}$

$\Rightarrow d(p-q-1+1)=\frac{2(aq-bp)}{pq}$

$\Rightarrow d(p-q)=\frac{2(aq-bp)}{pq}$

$\Rightarrow d=\frac{2(aq-bp)}{pq(p-q)}$

Subtracting equation (3) from (2), we have

$\Rightarrow (q-1)d-(r-1)d=\frac{2b}{q}-\frac{2c}{r}$

$\Rightarrow d(q-r-1+1)=\frac{2(br-cq)}{qr}$

$\Rightarrow d(q-r)=\frac{2(br-qc)}{qr}$

$\Rightarrow d=\frac{2(br-qc)}{qr(q-r)}$

Equating values of d, we have

$\Rightarrow d=\frac{2(aq-bp)}{pq(p-q)}$$=\frac{2(br-qc)}{qr(q-r)}$

$\Rightarrow \frac{2(aq-bp)}{pq(p-q)}$$=\frac{2(br-qc)}{qr(q-r)}$

$\Rightarrow \, \, (aq-bp)qr(q-r)=(br-qc)pq(p-q)$

$\Rightarrow \, \, (aq-bp)r(q-r)=(br-qc)p(p-q)$

$\Rightarrow \, \, (aqr-bpr)(q-r)=(bpr-pqc)(p-q)$

Dividing both sides from pqr, we get

$\Rightarrow \, \, (\frac{a}{p}-\frac{b}{q})(q-r)=(\frac{b}{q}-\frac{c}{r})(p-q)$

$\Rightarrow \, \, \frac{a}{p}(q-r)-\frac{b}{q}(q-r+p-q)+\frac{c}{r}(p-q)=0$

$\Rightarrow \, \, \frac{a}{p}(q-r)-\frac{b}{q}(p-r)+\frac{c}{r}(p-q)=0$

$\Rightarrow \, \, \frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)=0$

Hence, the given result is proved.

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