Question

If $a$ is the A.M. of two numbers $b$ and $c$, and ${G_1}$, ${G_2}$ are the two G.M.’s between them, then prove that $G_1^3 + G_2^3 = 2abc$.

Answer 1

Here, we need to prove the given equation. The two numbers, and the G.M.s between them are in G.P. We will use this, and the formula for ${n^{{\rm{th}}}}$ term of a G.P. to find the value of the two G.M.s. Then, we will put the values in the left hand side of the given expression. Finally, we will use the arithmetic mean formula to simplify the equation and prove the required equation.
Formula Used: We will use the following formulas:
1. The arithmetic mean of two numbers $a$ and $b$ is given by the formula $A.M. = \dfrac{{a + b}}{2}$.
2. The geometric mean of two numbers $a$ and $b$ is given by the formula $G.M. = \sqrt {ab}$. Here, $a$, $\sqrt {ab}$, $b$, forms a G.P.
3. The ${n^{{\rm{th}}}}$ term of a G.P. is given by the formula ${a_n} = a{r^{n - 1}}$, where $a$ is the first term and $r$ is the common ratio.

Complete step-by-step answer:
We will use the formula for A.M. and G.M. to prove that $G_1^3 + G_2^3 = 2abc$.
The arithmetic mean of two numbers $a$ and $b$ is given by the formula $A.M. = \dfrac{{a + b}}{2}$.
Therefore, we get the arithmetic mean of two numbers $b$ and $c$ as $A.M. = \dfrac{{b + c}}{2}$.
It is given that the arithmetic mean of $b$ and $c$ is $a$.
Therefore, we get
$\Rightarrow a = \dfrac{{b + c}}{2}$
Multiplying both sides by 2, we get
$\begin{array}{l} \Rightarrow a \times 2 = \left( {\dfrac{{b + c}}{2}} \right) \times 2\\\ \Rightarrow 2a = b + c\end{array}$
Next, the geometric mean of two numbers $a$ and $b$ is given by the formula $G.M. = \sqrt {ab}$. Here, $a$, $\sqrt {ab}$, $b$, forms a G.P.
Since ${G_1}$, ${G_2}$ are the two G.M.s between the numbers $b$ and $c$, therefore $b$, ${G_1}$, ${G_2}$, and $c$, are in G.P.
The ${n^{{\rm{th}}}}$ term of a G.P. is given by the formula ${a_n} = a{r^{n - 1}}$, where $a$ is the first term and $r$ is the common ratio.
The term $c$ is the 4th term of the G.P.
Therefore, substituting $n = 4$ and $b$ as the first term in the formula, we get
$\begin{array}{l} \Rightarrow {a_4} = b{r^{4 - 1}}\\\ \Rightarrow c = b{r^3}\end{array}$
Dividing both sides of the equation by $b$, we get
$\Rightarrow \dfrac{c}{b} = {r^3}$
Taking the cube roots of both sides, we get
$\Rightarrow r = {\left( {\dfrac{c}{b}} \right)^{\dfrac{1}{3}}}$
The term ${G_1}$ is the 2nd term of the G.P.
Therefore, substituting $n = 2$ and $b$ as the first term in the formula, we get
$\begin{array}{l} \Rightarrow {a_2} = b{r^{2 - 1}}\\\ \Rightarrow {G_1} = b{r^1}\\\ \Rightarrow {G_1} = br\end{array}$
Substituting $r = {\left( {\dfrac{c}{b}} \right)^{\dfrac{1}{3}}}$ in the expression, we get
$\Rightarrow {G_1} = b{\left( {\dfrac{c}{b}} \right)^{\dfrac{1}{3}}}$
Taking the cube of both sides, we get
$\begin{array}{l} \Rightarrow {\left[ {{G_1}} \right]^3} = {\left[ {b{{\left( {\dfrac{c}{b}} \right)}^{\dfrac{1}{3}}}} \right]^3}\\\ \Rightarrow G_1^3 = {b^3}{\left( {\dfrac{c}{b}} \right)^{\dfrac{1}{3} \times 3}}\end{array}$
Simplifying the expression, we get
$\begin{array}{l} \Rightarrow G_1^3 = {b^3}{\left( {\dfrac{c}{b}} \right)^1}\\\ \Rightarrow G_1^3 = {b^3}\left( {\dfrac{c}{b}} \right)\\\ \Rightarrow G_1^3 = {b^2}c\end{array}$
The term ${G_2}$ is the 3rd term of the G.P.
Therefore, substituting $n = 3$ and $b$ as the first term in the formula, we get
$\begin{array}{l} \Rightarrow {a_3} = b{r^{3 - 1}}\\\ \Rightarrow {G_2} = b{r^2}\end{array}$
Substituting $r = {\left( {\dfrac{c}{b}} \right)^{\dfrac{1}{3}}}$ in the expression, we get
$\Rightarrow {G_2} = b{\left[ {{{\left( {\dfrac{c}{b}} \right)}^{\dfrac{1}{3}}}} \right]^2}$
Simplifying the expression, we get
$\begin{array}{l} \Rightarrow {G_2} = b{\left( {\dfrac{c}{b}} \right)^{\dfrac{1}{3} \times 2}}\\\ \Rightarrow {G_2} = b{\left( {\dfrac{c}{b}} \right)^{\dfrac{2}{3}}}\end{array}$
Taking the cube of both sides, we get
$\begin{array}{l} \Rightarrow {\left[ {{G_2}} \right]^3} = {\left[ {b{{\left( {\dfrac{c}{b}} \right)}^{\dfrac{2}{3}}}} \right]^3}\\\ \Rightarrow G_2^3 = {b^3}{\left( {\dfrac{c}{b}} \right)^{\dfrac{2}{3} \times 3}}\end{array}$
Simplifying the expression, we get
$\begin{array}{l} \Rightarrow G_2^3 = {b^3}{\left( {\dfrac{c}{b}} \right)^2}\\\ \Rightarrow G_2^3 = {b^3}\left( {\dfrac{{{c^2}}}{{{b^2}}}} \right)\\\ \Rightarrow G_2^3 = b{c^2}\end{array}$
Finally, we will simplify the left hand side of the equation $G_1^3 + G_2^3 = 2abc$.
Substituting $G_1^3 = {b^2}c$ and $G_2^3 = b{c^2}$ in the expression $G_1^3 + G_2^3$, we get
$\Rightarrow G_1^3 + G_2^3 = {b^2}c + b{c^2}$
Factoring out the common term $bc$, we get
$\Rightarrow G_1^3 + G_2^3 = bc\left( {b + c} \right)$
Substituting $b + c = 2a$ in the expression, we get
$\Rightarrow G_1^3 + G_2^3 = bc\left( {2a} \right)$
Multiplying the terms, we get
$\Rightarrow G_1^3 + G_2^3 = 2abc$
$\therefore$ We have proved that $G_1^3 + G_2^3 = 2abc$.

Note: We used the geometric mean of two numbers $a$ and $b$. Since ${G_1}$, ${G_2}$ are the two G.M.s between the numbers $b$ and $c$, therefore $b$, ${G_1}$, ${G_2}$, and $c$, are in G.P. All the G.M.s between two numbers, when placed in sequence between the two numbers, form a geometric progression. This is true for any number of geometric means.