The pth, qth and rth terms of an A.P. are a, b, c respectively.
Question:

The $p^{\text {th }}, q^{\text {th }}$ and $t^{\text {th }}$ terms of an A.P. are $a, b, c$ respectively. Show that $(q-r) a+(r-p) b+(p-q) c=0$

Solution:

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

The $n^{\text {th }}$ term of an A.P. is given by, $a_{n}=t+(n-1) d$

Therefore,

$a_{p}=t+(p-1) d=a \ldots$

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

$a_{r}=t+(r-1) d=c \ldots(3)$

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

$(p-1-q+1) d=a-b$

$\Rightarrow(p-q) d=a-b$

$\therefore \mathrm{d}=\frac{\mathrm{a}-\mathrm{b}}{\mathrm{p}-\mathrm{q}}$ $\ldots(4)$

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

$(q-1-r+1) d=b-c$

$\Rightarrow(q-r) d=b-c$

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

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

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

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

$\Rightarrow a q-b q-a r+b r=b p-b q-c p+c q$

$\Rightarrow \mathrm{bp}-\mathrm{cp}+\mathrm{cq}-\mathrm{aq}+\mathrm{ar}-\mathrm{br}=0$

$\Rightarrow(-a q+a r)+(b p-b r)+(-c p+c q)=0 \quad$ (By rearranging terms)

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

 

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

Thus, the given result is proved.

 

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