Question

If $G$ is the geometric mean of $x$ and$y$ , then $\dfrac{1}{{{G}^{2}}-{{x}^{2}}}+\dfrac{1}{{{G}^{2}}-{{y}^{2}}}=$
1)${{G}^{2}}$
2)$\dfrac{1}{{{G}^{2}}}$
3) $\dfrac{2}{{{G}^{2}}}$
4) $\dfrac{3}{{{G}^{2}}}$

Answer 1

In this question we will first use a formula of geometric mean which is equal to the square root of two given numbers which are in geometric progression. Then we will substitute the value of the geometric mean in our given equation. After that we will simplify the given equation to get our required answer.

Complete step-by-step solution:
A sequence is an arrangement of any set of numbers in a particular order. We mainly know about two sequences or we can say that progression.
Arithmetic progression
Geometric progression
First we will learn some basic concepts of Arithmetic Progression.
In arithmetic progression we can add or subtract a particular number to create progression. That particular number is called common difference.
First term of a progression is denoted by $a$ and we represent the terms of arithmetic progression by ${{a}_{1}},{{a}_{2}},{{a}_{3}},{{a}_{4}},{{a}_{5}},{{a}_{6,}}........$
Common difference is denoted as $d$ . We find the value of $d$ by subtracting two successive terms of progression.
$d={{a}_{2}}-{{a}_{1}}={{a}_{3}}-{{a}_{2}}={{a}_{n}}-{{a}_{n-1}}$ .
${{n}^{th}}$Term is also denoted as ${{a}_{n}}$ .
The formula for ${{n}^{th}}$term of arithmetic progression is ${{a}_{n}}=a+\left( n-1 \right)d$
Where,
$a$ is first term
$d$ is a common difference of arithmetic progression.
In arithmetic progression we find arithmetic mean (A)
$A=\dfrac{1}{n}\sum\limits_{i=1}^{n}{{{a}_{i}}}$
Now we will learn some basic concepts of Geometric Progression.
In geometric progression (G.P) we multiply or divide a particular number to create progression. That particular number is called the common ratio .
First term of a progression is denoted by $a$ and we represent the terms of geometric progression by ${{a}_{1}},{{a}_{2}},{{a}_{3}},{{a}_{4}},{{a}_{5}},{{a}_{6,}}........$
Common ratio is denoted as $r$ . We find the value of $r$ by dividing two successive terms of progression.
$r=\dfrac{{{a}_{2}}}{{{a}_{1}}}=\dfrac{{{a}_{3}}}{{{a}_{2}}}=\dfrac{{{a}_{n}}}{{{a}_{n-1}}}$ .
${{n}^{th}}$Term is denoted as ${{a}_{n}}$ .
The formula for ${{n}^{th}}$ term of geometric progression is
${{a}_{n}}=a{{r}^{n-1}}$
Where,
$a$ is first term
$r$ is a common ratio of geometric progression.
.Now we will continue to our question
$\dfrac{1}{{{G}^{2}}-{{x}^{2}}}+\dfrac{1}{{{G}^{2}}-{{y}^{2}}}$
$G$ is a geometric mean of $x$&$y$ . By above formula of geometric mean ,
$G=\sqrt{xy}$
Substituting the value of $G$ in given above equation
$\dfrac{1}{{{G}^{2}}-{{x}^{2}}}+\dfrac{1}{{{G}^{2}}-{{y}^{2}}}$
Now the substituting the value of $G$if above equation, we get
$\dfrac{1}{{{\left( \sqrt{xy} \right)}^{2}}-{{x}^{2}}}+\dfrac{1}{{{\left( \sqrt{xy} \right)}^{2}}-{{y}^{2}}}$
$\Rightarrow \dfrac{1}{xy-{{x}^{2}}}+\dfrac{1}{xy-{{y}^{2}}}$
$\Rightarrow \dfrac{1}{x\left( y-x \right)}+\dfrac{1}{y\left( x-y \right)}$
$\Rightarrow \dfrac{1}{x\left( y-x \right)}-\dfrac{1}{y\left( y-x \right)}$
$\begin{aligned} & \Rightarrow \dfrac{y-x}{xy\left( y-x \right)} \\\ & \Rightarrow \dfrac{1}{xy} \\\ & \Rightarrow \dfrac{1}{{{\left( \sqrt{xy} \right)}^{2}}} \\\ & \therefore \dfrac{1}{{{G}^{2}}} \\\ \end{aligned}$
$\therefore$ The required answer is $\dfrac{1}{{{G}^{2}}}$
Hence the Correct option is $\left( 2 \right)$.

Note: Generally mean denotes the average of given data .Here we have to know about three different means first is arithmetic mean and second geometric mean and third is Harmonic mean.
Arithmetic mean is calculated by just adding the two given numbers in and then divide it by two .