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

Question:

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

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

Solution:

Let a1 and d be the first term and the common difference of the A.P. respectively.

According to the given information,

$S_{p}=\frac{p}{2}\left[2 a_{1}+(p-1) d\right]=a$

$\Rightarrow 2 a_{1}+(p-1) d=\frac{2 a}{p}$ $\ldots(1)$

$S_{q}=\frac{q}{2}\left[2 a_{1}+(q-1) d\right]=b$

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

$S_{r}=\frac{r}{2}\left[2 a_{1}+(r-1) d\right]=c$

$\Rightarrow 2 a_{1}+(r-1) d=\frac{2 c}{r}$ (3)

Subtracting (2) from (1), we obtain

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

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

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

$\Rightarrow d=\frac{2(a q-b p)}{p q(p-q)}$ $\cdots \cdots(4)$

Subtracting (3) from (2), we obtain

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

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

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

$\Rightarrow d=\frac{2(b r-q c)}{q r(q-r)}$ $\ldots(5)$

Equating both the values of d obtained in (4) and (5), we obtain

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

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

$\Rightarrow r(q-r)(a q-b p)=p(p-q)(b r-q c)$

$\Rightarrow r(a q-b p)(q-r)=p(b r-q c)(p-q)$

$\Rightarrow(a q r-b p r)(q-r)=(b p r-c p q)(p-q)$

Dividing both sides by pqr, we obtain

$\left(\frac{a}{p}-\frac{b}{q}\right)(q-r)=\left(\frac{b}{q}-\frac{c}{r}\right)(p-q)$

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

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

Thus, the given result is proved.

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