If three unequal numbers p, q, r are in H.P. | vedclass

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(D) Given that $p, q, r$ are in $H.P.$,we have $q = \frac{2pr}{p+r}$.Let $K = \frac{pr}{p+r}$,then $q = 2K$ and $pr = K(p+r)$.Since $p^2, q^2, r^2$ ar

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$1 \mp \sqrt{3} : - 2 : 1 \pm \sqrt{3}$

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(D) Given that $p, q, r$ are in $H.P.$,we have $q = \frac{2pr}{p+r}$.Let $K = \frac{pr}{p+r}$,then $q = 2K$ and $pr = K(p+r)$.Since $p^2, q^2, r^2$ are in $A.P.$,we have $2q^2 = p^2 + r^2$.Substituting $q = 2K$,we get $2(4K^2) = p^2 + r^2$,so $8K^2 = p^2 + r^2$.We know $(p+r)^2 = p^2 + r^2 + 2pr = 8K^2 + 2K(p+r)$.Let $S = p+r$. Then $S^2 - 2KS - 8K^2 = 0$.Solving for $S$,$(S - 4K)(S + 2K) = 0$,so $S = 4K$ or $S = -2K$.If $S = 4K$,then $(p-r)^2 = (p+r)^2 - 4pr = 16K^2 - 4(4K^2) = 0$,implying $p=r$,which contradicts the condition that numbers are unequal.If $S = -2K$,then $pr = K(-2K) = -2K^2$.Then $(p-r)^2 = (p+r)^2 - 4pr = (-2K)^2 - 4(-2K^2) = 4K^2 + 8K^2 = 12K^2$.Thus $p-r = \pm 2\sqrt{3}K$.Solving $p+r = -2K$ and $p-r = \pm 2\sqrt{3}K$,we get $p = (-1 \pm \sqrt{3})K$ and $r = (-1 \mp \sqrt{3})K$.Thus $p:q:r = (-1 \pm \sqrt{3})K : 2K : (-1 \mp \sqrt{3})K = 1 \mp \sqrt{3} : -2 : 1 \pm \sqrt{3}$.

If three unequal numbers $p, q, r$ are in $H.P.$ and their squares are in $A.P.$,then the ratio $p:q:r$ is

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