If pth, qth and rth terms of an A.P. are in G.P. whose common ratio is k, then the root of equation
$\left(q-r\right)x^2+\left(r-p\right)x+\left(p-q\right)=0$ other than unity is :
(1) k ²               (2) k                 (3) 2k                (4) None of these

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{pth}, \mathrm{qth} \mathrm{and} \mathrm{rth} \mathrm{of} \mathrm{A}.\mathrm{P}. \mathrm{are} \mathrm{in} \mathrm{G}.\mathrm{P}. \\ \mathrm{If} \mathrm{a} \mathrm{is} \mathrm{first} \mathrm{term} \mathrm{and} \mathrm{d} \mathrm{is} \mathrm{common} \mathrm{difference}, \mathrm{then} \\ {\mathrm{T}}_{\mathrm{p}}=\mathrm{a}+\left(\mathrm{p}-1\right)\mathrm{d} \\ {\mathrm{T}}_{\mathrm{q}}=\mathrm{a}+\left(\mathrm{q}-1\right)\mathrm{d} \\ {\mathrm{T}}_{\mathrm{r}}=\mathrm{a}+\left(\mathrm{r}-1\right)\mathrm{d} \\ \mathrm{As} \mathrm{they} \mathrm{are} \mathrm{in} \mathrm{G}.\mathrm{P}. \\ \frac{{\mathrm{T}}_{\mathrm{q}}}{{\mathrm{T}}_{\mathrm{p}}}=\frac{{\mathrm{T}}_{\mathrm{r}}}{{\mathrm{T}}_{\mathrm{q}}} \\ \Rightarrow \frac{\mathrm{a}+\left(\mathrm{q}-1\right)\mathrm{d}}{\mathrm{a}+\left(\mathrm{p}-1\right)\mathrm{d}}=\frac{\mathrm{a}+\left(\mathrm{r}-1\right)\mathrm{d}}{\mathrm{a}+\left(\mathrm{q}-1\right)\mathrm{d}}=\mathrm{k} \left(\mathrm{Common} \mathrm{ratio}\right) \\ \Rightarrow \frac{\mathrm{a}+\left(\mathrm{q}-1\right)\mathrm{d}}{\mathrm{a}+\left(\mathrm{p}-1\right)\mathrm{d}}-1=\frac{\mathrm{a}+\left(\mathrm{r}-1\right)\mathrm{d}}{\mathrm{a}+\left(\mathrm{q}-1\right)\mathrm{d}}-1 \\ \Rightarrow \frac{\mathrm{a}+\left(\mathrm{q}-1\right)\mathrm{d}-\mathrm{a}-\left(\mathrm{p}-1\right)\mathrm{d}}{\mathrm{a}+\left(\mathrm{p}-1\right)\mathrm{d}}=\frac{\mathrm{a}+\left(\mathrm{r}-1\right)\mathrm{d}-\mathrm{a}-\left(\mathrm{q}-1\right)\mathrm{d}}{\mathrm{a}+\left(\mathrm{q}-1\right)\mathrm{d}} \\ \Rightarrow \frac{\mathrm{d}\left(\mathrm{q}-\mathrm{p}\right)}{\mathrm{a}+\left(\mathrm{p}-1\right)\mathrm{d}}=\frac{\mathrm{d}\left(\mathrm{r}-\mathrm{q}\right)}{\mathrm{a}+\left(\mathrm{q}-1\right)\mathrm{d}} \\ \Rightarrow \frac{\mathrm{a}+\left(\mathrm{q}-1\right)\mathrm{d}}{\mathrm{a}+\left(\mathrm{p}-1\right)\mathrm{d}}=\frac{\mathrm{r}-\mathrm{q}}{\mathrm{q}-\mathrm{p}}=\frac{\mathrm{q}-\mathrm{r}}{\mathrm{p}-\mathrm{q}} \\ \Rightarrow \mathrm{k}=\frac{\mathrm{q}-\mathrm{r}}{\mathrm{p}-\mathrm{q}} \\ \mathrm{Now} \\ \left(\mathrm{q}-\mathrm{r}\right){\mathrm{x}}^2+\left(\mathrm{r}-\mathrm{p}\right)\mathrm{x}+\left(\mathrm{p}-\mathrm{q}\right)=0 \\ \mathrm{If} \mathrm{we} \mathrm{put} \mathrm{x}=1 \mathrm{in} \mathrm{L}.\mathrm{H}.\mathrm{S}> \mathrm{it} \mathrm{becomes} \mathrm{q}-\mathrm{r}+\mathrm{r}-\mathrm{p}+\mathrm{p}-\mathrm{q}=0 , \mathrm{hence} \\ 1 \mathrm{is} \mathrm{root} \mathrm{of} \mathrm{above} \mathrm{equation}, \mathrm{let} \mathrm{other} \mathrm{root} \mathrm{be} \mathrm{\alpha } \\ \Rightarrow \mathrm{Product} \mathrm{of} \mathrm{roots}=1\times \mathrm{\alpha }=\frac{\mathrm{p}-\mathrm{q}}{\mathrm{q}-\mathrm{r}} \\ \Rightarrow \mathrm{\alpha }=\frac{\mathrm{p}-\mathrm{q}}{\mathrm{q}-\mathrm{r}} \\ \Rightarrow \mathrm{\alpha }=\frac{1}{\mathrm{k}} \left[\mathrm{As} \mathrm{k}=\frac{\mathrm{q}-\mathrm{r}}{\mathrm{p}-\mathrm{q}}\right] \\ \mathrm{Option}\left(\mathrm{d}\right) \end{gathered}$$

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