CMAT 2014 Solved Question Paper

(c) Since, roots of the equation are zero.
We know that, ${\mathit{b}}^2$ - 4ac = 0
From equation,
${\mathit{q}}^2{\left(\mathit{r}-\mathit{p}\right)}^2$ - 4pr(q - r )(p -q ) = 0
${\mathit{q}}^2{\left(\mathit{r}-\mathit{p}\right)}^2$ - 4pr(pq - pr - ${\mathit{q}}^2$ +qr) = 0
${\mathit{q}}^2{\mathit{r}}^2+{\mathit{q}}^2{\mathit{p}}^2$ - $2\mathit{p}{\mathit{q}}^2\mathit{r}-4{\mathit{p}}^2\mathit{q}\mathit{r}$ + $4{\mathit{p}}^2{\mathit{r}}^2+4\mathit{p}{\mathit{q}}^2\mathit{r}$ - $4\mathit{p}\mathit{q}{\mathit{r}}^2$ = 0
${\mathit{q}}^2{\left(\mathit{p}+\mathit{r}\right)}^2$ + $4{\mathit{p}}^2{\mathit{r}}^2$ - 4pqr(p + r) = 0
${\left[\mathit{q}\left(\mathit{p}+\mathit{r}\right)-2\mathit{p}\mathit{r}\right]}^2$ =0
q(p + r) = 2pr
∴ q = $\frac{2\mathit{p}\mathit{r}}{\mathit{p}+\mathit{r}}$
Hence, p, q and r are in HP.