Question

If ${G_1}{\text{ and }}{G_2}$ are two geometric mean and A is the arithmetic mean inserted between two numbers, then the value of $\dfrac{{{G_1}^2}}{{{G_2}}} + \dfrac{{{G_2}^2}}{{{G_1}}}$ is
A) 4A
B) A
C) 2A
D) 3A

Answer 1

We have two geometric mean ${G_1}{\text{ and }}{G_2}$ and arithmetic mean A inserted between two numbers. We have to find the value of $\dfrac{{{G_1}^2}}{{{G_2}}} + \dfrac{{{G_2}^2}}{{{G_1}}}$the term.
Arithmetic mean of two numbers a and b is $\dfrac{{a + b}}{2}$
Geometric mean between these two numbers a, b is ${(a.b)^{\dfrac{1}{2}}}$
Formula used- nth term of geometric mean is $a{r^{n - 1}}$
Here is the initial term of the geometric mean.
n is the number of terms in geometric mean.
r is a comment ratio of the geometric mean.

Complete step by step answer:
Let us consider the two numbers as a, b.
By formula arithmetic mean of two numbers a and b is $\dfrac{{a + b}}{2}$
From the given question we have, the Arithmetic mean of two numbers a and b is A.
Then
$A = \dfrac{{a + b}}{2}$
On solving the above equation we have,
$a + b = 2A......(1)$
Let us mark the above equation as equation (1)
Also it is given that, the geometric mean between these two numbers a, b are${G_1}{\text{ and }}{G_2}$.
That is $a,{G_1},{G_2},b$, is the given sequence derived from the given question.
Let us consider r to be the common ratio, then the sequence following geometric progression be like $a,ar,a{r^2},a{r^3}$
By using the formula to find the nth term of geometric mean, we get,
Let us compare the derived sequence with the general form of geometric progression,
Then we get,
$b = a{r^{4 - 1}}$
On solving the above equation we have,
$b = a{r^3}$
Let us again solve the equation to find the value of r
$\dfrac{b}{a} = {r^3}$
$r = \sqrt[3]{{\dfrac{b}{a}}}$
Now let us substitute the value of r in the general term and compare with the derived series,
Thus ${G_1} = ar = a\sqrt[3]{{\dfrac{b}{a}}} = a.{(\dfrac{b}{a})^{\dfrac{1}{3}}} = {a^{1 - \dfrac{1}{3}}}{b^{\dfrac{1}{3}}} = {a^{\dfrac{2}{3}}}{b^{\dfrac{1}{3}}}$
Also, ${G_2} = a{r^2} = a{(\sqrt[3]{{\dfrac{b}{a}}})^2} = a.{(\dfrac{b}{a})^{\dfrac{2}{3}}} = {a^{1 - \dfrac{2}{3}}}{b^{\dfrac{2}{3}}} = {a^{\dfrac{1}{3}}}{b^{\dfrac{2}{3}}}$
Hence we have found the value of ${G_1}{\text{ and }}{G_2}$
We are given that to find the value of the following equation,
$\dfrac{{{G_1}^2}}{{{G_2}}} + \dfrac{{{G_2}^2}}{{{G_1}}}$
On solving we get,
$\dfrac{{{G_1}^3 + {G_2}^3}}{{{G_1}{G_2}}}......(2)$
Let us consider this as equation (2)
Let us substitute the value of ${G_1}{\text{ and }}{G_2}$in the following then we have,
${G_1}^3 = {({a^{\dfrac{2}{3}}}{b^{\dfrac{1}{3}}})^3} = {a^2}b$
${G_2}^3 = {({a^{\dfrac{1}{3}}}{b^{\dfrac{2}{3}}})^3} = a{b^2}$
${G_1}{G_2} = {a^{\dfrac{2}{3}}}{b^{\dfrac{1}{3}}} \times {a^{\dfrac{1}{3}}}{b^{\dfrac{2}{3}}} = {a^{\dfrac{2}{3} + \dfrac{1}{3}}}{b^{\dfrac{1}{3} + \dfrac{2}{3}}} = {a^{\dfrac{3}{3}}}{b^{\dfrac{3}{3}}} = ab$
Now let us substitute the value found above in equation (2), we get
$\dfrac{{{G_1}^2}}{{{G_2}}} + \dfrac{{{G_2}^2}}{{{G_1}}} = \dfrac{{{a^2}b + a{b^2}}}{{ab}}$
Let us solve the fractions in the above equation, then we get
$\dfrac{{{G_1}^2}}{{{G_2}}} + \dfrac{{{G_2}^2}}{{{G_1}}} = \dfrac{{ab(a + b)}}{{ab}}$
$\dfrac{{{G_1}^2}}{{{G_2}}} + \dfrac{{{G_2}^2}}{{{G_1}}} = a + b$
From equation (1) we have $a + b = 2A$, substituting in the above equation we get
$\dfrac{{{G_1}^2}}{{{G_2}}} + \dfrac{{{G_2}^2}}{{{G_1}}} = 2A$

Hence, $2A$ is in option (C). It is the correct option.

Note:
We have used the formula ${a^m} \times {a^n} = {a^{m + n}}$ to find the values of ${G_1}{\text{ and }}{G_2}$. Which is required the most to solve the ${G_1}{\text{ and }}{G_2}$.
Geometric mean: The geometric mean is the average of a set of products, the calculation of which is commonly used to determine the performance results of an investment.
Arithmetic mean: The arithmetic mean or simply called average is the ratio of all observations to the total number of observations.