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

Hint: 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.