Question

If AM and GM of $x$ and $y$are in the ratio $p:q$, then $x:y$is
1). $p-\sqrt{{{p}^{2}}+{{q}^{2}}}:p+\sqrt{{{p}^{2}}+{{q}^{2}}}$
2). $p+\sqrt{{{p}^{2}}-{{q}^{2}}}:p-\sqrt{{{p}^{2}}-{{q}^{2}}}$
3). $p:q$
4). None of these.

Answer 1

To get the solution of this question firstly we have to understand the concept used in this question and the concepts are arithmetic mean and geometric mean. After applying these concepts, use the componendo and dividendo rule to further simplify the expression. Using these you can get your answer.

Complete step-by-step solution:
To solve this question firstly we have to understand the concepts used in this question and they are Arithmetic Mean and Geometric Mean. There are various ways to calculate the central value and they are mean, median and mode. But mean is generally used to take the average of all the numbers of the sequence
Arithmetic Mean is calculated by taking the summation of all the terms present in the series divided by the total number of terms in the series.
Let the given series is ${{x}_{1}},{{x}_{2}},{{x}_{3}},{{x}_{4}},.....,{{x}_{n}}$ then,
$AM=\dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}+....+{{x}_{n}}}{n}$
Geometric Mean is calculated by taking the ${{n}^{th}}$ root of the product of all the terms given in the series. Mathematically it is represented as
Let the given series is ${{x}_{1}},{{x}_{2}},{{x}_{3}},{{x}_{4}},.....,{{x}_{n}}$ then,
$GM=\sqrt[n]{{{x}_{1}}.{{x}_{2}}.{{x}_{3}}.{{x}_{4}}......{{x}_{n}}}$
Now let us apply this concept in the given question to get the answer.
According to the definition, AM of $x$ and $y$is
$AM=\dfrac{x+y}{2}$
Similarly according to the definition, GM of $x$ and $y$ is
$GM=\sqrt{xy}$
Now it is given that the ratio of AM and GM is $p:q$. Mathematically we can represent as
$\dfrac{AM}{GM}=\dfrac{p}{q}$
Substituting the value of AM and GM, we get
$\dfrac{x+y}{2\sqrt{xy}}=\dfrac{p}{q}$
There is a theorem known as componendo and dividendo which makes the calculation of ratio and proportion easier. To solve this ratio apply the same theorem and we will get
$\Rightarrow \dfrac{x+y+2\sqrt{xy}}{x+y-2\sqrt{xy}}=\dfrac{p+q}{p-q}$
$\Rightarrow \dfrac{{{(\sqrt{x}+\sqrt{y})}^{2}}}{{{(\sqrt{x}-\sqrt{y})}^{2}}}=\dfrac{p+q}{p-q}$
Now taking square root on both the sides, we get
$\Rightarrow \dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\dfrac{\sqrt{p+q}}{\sqrt{p-q}}$
Again applying componendo and dividendo rule, we will get
$\Rightarrow \dfrac{\sqrt{x}+\sqrt{y}+\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}-\sqrt{x}+\sqrt{y}}=\dfrac{\sqrt{p+q}+\sqrt{p-q}}{\sqrt{p+q}-\sqrt{p-q}}$

& \Rightarrow \dfrac{2\sqrt{x}}{2\sqrt{y}}=\dfrac{\sqrt{p+q}+\sqrt{p-q}}{\sqrt{p+q}-\sqrt{p-q}} \\\ & \Rightarrow \dfrac{\sqrt{x}}{\sqrt{y}}=\dfrac{\sqrt{p+q}+\sqrt{p-q}}{\sqrt{p+q}-\sqrt{p-q}} \\\ \end{aligned}$$ Squaring on both the sides, we will get $$\Rightarrow \dfrac{x}{y}=\dfrac{p+q+p-q+2\sqrt{p+q}\sqrt{p-q}}{p+q+p-q-2\sqrt{p+q}\sqrt{p-q}}$$ $$\Rightarrow \dfrac{x}{y}=\dfrac{2p+2\sqrt{p+q}\sqrt{p-q}}{2p-2\sqrt{p+q}\sqrt{p-q}}$$ $$\Rightarrow \dfrac{x}{y}=\dfrac{2p+2\sqrt{{{p}^{2}}-{{q}^{2}}}}{2p-2\sqrt{{{p}^{2}}-{{q}^{2}}}}$$ Taking $$2$$ common from both numerator and denominator, after cancelling out $$2$$, we will get $$\Rightarrow \dfrac{x}{y}=\dfrac{p+\sqrt{{{p}^{2}}-{{q}^{2}}}}{p-\sqrt{{{p}^{2}}-{{q}^{2}}}}$$ From the above expression we can say that the ratio of $$x:y$$ is $$p+\sqrt{{{p}^{2}}-{{q}^{2}}}:p-\sqrt{{{p}^{2}}-{{q}^{2}}}$$ **Hence we can conclude that the option $$(2)$$ is the correct one.** **Note:** Geometric mean is always less than the arithmetic mean for the series consisting of the same numbers. Geometric mean is only applicable for positive series i.e. all the terms of the series should be positive and there should be no zero value in the series. GM is used in the finance department to know the growth rates.