If the arithmetic mean between p and q (p q | vedclass

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(D) The arithmetic mean $(AM)$ of $p$ and $q$ is $\frac{p+q}{2}$.The geometric mean $(GM)$ of $p$ and $q$ is $\sqrt{pq}$.Given that $AM = 2 	imes GM$

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$(7 + 4\sqrt{3}) : 1$

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(D) The arithmetic mean $(AM)$ of $p$ and $q$ is $\frac{p+q}{2}$.The geometric mean $(GM)$ of $p$ and $q$ is $\sqrt{pq}$.Given that $AM = 2 	imes GM$,we have $\frac{p+q}{2} = 2\sqrt{pq}$.This implies $p+q = 4\sqrt{pq}$.Squaring both sides,we get $(p+q)^2 = 16pq$,which simplifies to $p^2 + 2pq + q^2 = 16pq$,or $p^2 - 14pq + q^2 = 0$.Dividing by $q^2$,we get $(\frac{p}{q})^2 - 14(\frac{p}{q}) + 1 = 0$.Let $x = \frac{p}{q}$. Then $x^2 - 14x + 1 = 0$.Using the quadratic formula,$x = \frac{14 \pm \sqrt{196 - 4}}{2} = \frac{14 \pm \sqrt{192}}{2} = \frac{14 \pm 8\sqrt{3}}{2} = 7 \pm 4\sqrt{3}$.Since $p > q$,$x > 1$,so $x = 7 + 4\sqrt{3} = \frac{7 + 4\sqrt{3}}{1}$.Note that $(2 + \sqrt{3})^2 = 4 + 3 + 4\sqrt{3} = 7 + 4\sqrt{3}$.Thus,the ratio $p:q$ is $(7 + 4\sqrt{3}) : 1$.

If the arithmetic mean between $p$ and $q$ $(p > q)$ is twice the geometric mean,then $p : q = .......$

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