If p, q, r are in A.P. and are positive,the r | vedclass

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(A) Given that $p, q, r$ are in $A.P.$ and are positive.Since they are in $A.P.$,we have $q = \frac{p + r}{2}$ ......$(i)$For the quadratic equation $

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$\left| \frac{r}{p} - 7 \right| \ge 4\sqrt{3}$

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(A) Given that $p, q, r$ are in $A.P.$ and are positive.Since they are in $A.P.$,we have $q = \frac{p + r}{2}$ ......$(i)$For the quadratic equation $px^2 + qx + r = 0$ to have real roots,the discriminant $D$ must be greater than or equal to $0$.$D = q^2 - 4pr \ge 0$Substituting $q = \frac{p + r}{2}$ into the inequality:$\left( \frac{p + r}{2} \right)^2 - 4pr \ge 0$$\frac{p^2 + 2pr + r^2}{4} - 4pr \ge 0$$p^2 + 2pr + r^2 - 16pr \ge 0$$p^2 - 14pr + r^2 \ge 0$Dividing by $p^2$ (since $p > 0$):$1 - 14\left( \frac{r}{p} \right) + \left( \frac{r}{p} \right)^2 \ge 0$Let $x = \frac{r}{p}$. Then $x^2 - 14x + 1 \ge 0$.Completing the square: $(x - 7)^2 - 49 + 1 \ge 0$$(x - 7)^2 \ge 48$$|x - 7| \ge \sqrt{48} = 4\sqrt{3}$Thus,$\left| \frac{r}{p} - 7 \right| \ge 4\sqrt{3}$.

If $p, q, r$ are in $A.P.$ and are positive,the roots of the quadratic equation $px^2 + qx + r = 0$ are all real for

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