Question

Let x be the arithmetic mean and y, z be the two geometric mean between any two positive numbers, then $\dfrac{{{y}^{3}}+{{z}^{3}}}{xyz}=\_\_\_\_\_$

Answer 1

Hint: Arithmetic mean of two positive numbers a, b is $x=\dfrac{a+b}{2}$. Consider a, y, z, b to be the geometric progression where y and z are geometric mean between a and b. Using this progression find the common ratio, $r={{\left( \dfrac{{{a}_{n}}}{{{a}_{1}}} \right)}^{\dfrac{1}{n-1}}}$ where ${{a}_{1}}$ is the first term, ${{a}_{n}}$ is the last term and $n$ is the total number of terms. Find y and z using $y=ar$ and $z=a{{r}^{2}}$. Substitute the values of x, y and z in $\dfrac{{{y}^{3}}+{{z}^{3}}}{xyz}$.

Complete step by step answer:
We are given in question that x is the arithmetic mean and y, z be two geometric mean between any two positive numbers. So let us consider two positive numbers to be a and b.
Then the arithmetic mean of these two numbers will be equal to the sum of a and b divided by 2, where 2 is the total number of terms.
Thus, arithmetic mean, $x=\dfrac{a+b}{2}$
Let y and z are two geometric means between a and b. Since y and z are geometric means they will surely lie between a and b. So a, y, z, b will be in geometric progression (G.P), now we will be calculating common ratio(r).
We know that if ${{a}_{1}},{{a}_{1}}r,{{a}_{1}}{{r}^{2}},.....,{{a}_{1}}{{r}^{n-1}},{{a}_{1}}{{r}^{n}}$ be a sequence of G.P then ${{n}^{th}}$ term sequence is given by ${{a}_{n}}={{a}_{1}}{{r}^{n-1}}$ where $r$ is the common ratio, ${{a}_{1}}$ is the first term, $n$ is the number of terms.
Therefore, ${{a}_{n}}={{a}_{1}}{{r}^{n-1}}$ can be written as
$\dfrac{{{a}_{n}}}{{{a}_{1}}}={{r}^{n-1}}$
Now we will take ${{\left( \dfrac{1}{n-1} \right)}^{th}}$ power on both sides we get,
${{\left( \dfrac{{{a}_{n}}}{{{a}_{1}}} \right)}^{\dfrac{1}{n-1}}}=r.........(1)$
We can see that the first term is a, the last term is b and the total number of terms is 4 in G.P a, y, z, b. Then common ratio, r can be calculated using equation (1), thus substituting ${{a}_{1}}=a$, ${{a}_{n}}=b$ and $n=4$ in equation (1) we get
$r={{\left( \dfrac{b}{a} \right)}^{\dfrac{1}{4-1}}}$
$r={{\left( \dfrac{b}{a} \right)}^{\dfrac{1}{3}}}$
In G.P a, y, z, b we see that y is the second term so $y=ar$ using G.P ${{a}_{1}},{{a}_{1}}r,{{a}_{1}}{{r}^{2}},.....,{{a}_{1}}{{r}^{n-1}},{{a}_{1}}{{r}^{n}}$. Thus in $y=ar$substituting the value of $r={{\left( \dfrac{b}{a} \right)}^{\dfrac{1}{3}}}$ we get,
$y=a\times {{\left( \dfrac{b}{a} \right)}^{\dfrac{1}{3}}}........(2)$
In G.P a, y, z, b we see that z is the third term of G.P so $z=a{{r}^{2}}$ using G.P ${{a}_{1}},{{a}_{1}}r,{{a}_{1}}{{r}^{2}},.....,{{a}_{1}}{{r}^{n-1}},{{a}_{1}}{{r}^{n}}$. Thus in $z=a{{r}^{2}}$ substituting the value of $r={{\left( \dfrac{b}{a} \right)}^{\dfrac{1}{3}}}$ we get,

$z=a\times {{\left( \dfrac{b}{a} \right)}^{\dfrac{2}{3}}}.........(3)$
Now putting the values of y and z from equation (2) and (3) in $\dfrac{{{y}^{3}}+{{z}^{3}}}{xyz}$we get,
$\dfrac{{{y}^{3}}+{{z}^{3}}}{xyz}=\dfrac{{{\left[ a\times {{\left( \dfrac{b}{a} \right)}^{\dfrac{1}{3}}} \right]}^{3}}+{{\left[ a\times {{\left( \dfrac{b}{a} \right)}^{\dfrac{2}{3}}} \right]}^{3}}}{\left( \dfrac{a+b}{2} \right)\left[ a\times {{\left( \dfrac{b}{a} \right)}^{\dfrac{1}{3}}} \right]\left[ a\times {{\left( \dfrac{b}{a} \right)}^{\dfrac{2}{3}}} \right]}$
In the numerator cubing both the terms and in the denominator multiplying the terms $\left[ a\times {{\left( \dfrac{b}{a} \right)}^{\dfrac{1}{3}}} \right]\left[ a\times {{\left( \dfrac{b}{a} \right)}^{\dfrac{2}{3}}} \right]$ we get,
$=\dfrac{{{\left[ {{a}^{3}}\times \left( \dfrac{b}{a} \right) \right]}^{{}}}+\left[ {{a}^{3}}\times {{\left( \dfrac{b}{a} \right)}^{2}} \right]}{\left( \dfrac{a+b}{2} \right)\left[ {{a}^{2}}\times \left( \dfrac{b}{a} \right) \right]}$
$=\dfrac{{{\left[ {{a}^{2}}b \right]}^{{}}}+\left[ a{{b}^{2}} \right]}{\left( \dfrac{a+b}{2} \right)\left[ ab \right]}$
Taking $ab$ common from the numerator we get $ab(a+b)$. The above term gets simplified to,
$=\dfrac{ab(a+b)}{\left( \dfrac{a+b}{2} \right)(ab)}$
In numerator and denominator $(ab)(a+b)$ are common so they get cancelled
$\begin{align}
  & =\dfrac{1}{\left( \dfrac{1}{2} \right)} \\
 & =2 \\
\end{align}$

Hence the value of $\dfrac{{{y}^{3}}+{{z}^{3}}}{xyz}$ is 2.

Note: If ${{a}_{1}},{{a}_{1}}r,{{a}_{1}}{{r}^{2}},.....,{{a}_{1}}{{r}^{n-1}},{{a}_{1}}{{r}^{n}}$ be a sequence of G.P then ${{n}^{th}}$ term sequence is given by ${{a}_{n}}={{a}_{1}}{{r}^{n-1}}$ where $r$ is the common ratio, ${{a}_{1}}$ is the first term, $n$ is the number of terms.
Therefore, ${{a}_{n}}={{a}_{1}}{{r}^{n-1}}$ can be written as $\dfrac{{{a}_{n}}}{{{a}_{1}}}={{r}^{n-1}}$ taking $\dfrac{1}{n-1}$power on both sides we get, common ratio as
$r={{\left( \dfrac{{{a}_{n}}}{{{a}_{1}}} \right)}^{\dfrac{1}{n-1}}}$. Care should be taken while evaluating $\dfrac{{{y}^{3}}+{{z}^{3}}}{xyz}=\dfrac{{{\left[ a\times {{\left( \dfrac{b}{a} \right)}^{\dfrac{1}{3}}} \right]}^{3}}+{{\left[ a\times {{\left( \dfrac{b}{a} \right)}^{\dfrac{2}{3}}} \right]}^{3}}}{\left( \dfrac{a+b}{2} \right)\left[ a\times {{\left( \dfrac{b}{a} \right)}^{\dfrac{1}{3}}} \right]\left[ a\times {{\left( \dfrac{b}{a} \right)}^{\dfrac{2}{3}}} \right]}$ proper brackets should be given so that confusion is not created.