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Hint: We will try to find out the common difference between ${{x}^{2}}+xy+{{y}^{2}}\And {{z}^{2}}+xz+{{x}^{2}}\And {{y}^{2}}+yz+{{z}^{2}}$ to prove they are in AP.

__Complete step by step answer:__

We all know that if there are three numbers in AP then they can be written as $\left( a-d \right),a$ and $\left( a+d \right)$. So, let us assume y=a, x=a-d and z=a+d.

We will now put x, y and z in the equation ${{x}^{2}}+xy+{{y}^{2}}$ .

${{\left( a-d \right)}^{2}}+\left( a-d \right)a+{{a}^{2}}$

Upon simplifying we get,

$={{a}^{2}}+{{d}^{2}}-2ad+{{a}^{2}}-ad+{{a}^{2}}$

$=3{{a}^{2}}+{{d}^{2}}-3ad.........\left( i \right)$

Similarly, we will put these values in ${{z}^{2}}+xz+{{x}^{2}}$ .

\[\begin{align}

& {{\left( a+d \right)}^{2}}+\left( a-d \right)\left( a+d \right)+{{\left( a-d \right)}^{2}} \\

& ={{a}^{2}}+{{d}^{2}}+2ad+{{a}^{2}}-{{d}^{2}}+{{a}^{2}}+{{d}^{2}}-2ad \\

& =3{{a}^{2}}+{{d}^{2}}..........\left( ii \right) \\

\end{align}\]

Now, we will put the values of x, y and z in the equation ${{y}^{2}}+yz+{{z}^{2}}$ .

\[\begin{align}

& {{a}^{2}}+a\left( a+d \right)+{{\left( a+d \right)}^{2}} \\

& ={{a}^{2}}+{{a}^{2}}+ad+{{a}^{2}}+{{d}^{2}}+2ad \\

& =3{{a}^{2}}+{{d}^{2}}+3ad..........\left( iii \right) \\

\end{align}\]

Now we will find the difference between equation (i) and equation (ii).

$\left( 3{{a}^{2}}+{{d}^{2}} \right)-\left( 3{{a}^{2}}+{{d}^{2}}-3ad \right)=3ad$

Similarly, we can find the difference between equation (ii) and equation (iii).

$\left( 3{{a}^{2}}+{{d}^{2}}+3ad \right)-\left( 3{{a}^{2}}+{{d}^{2}} \right)=3ad$

We can see from the above equations that the common difference between ${{x}^{2}}+xy+{{y}^{2}}\And {{z}^{2}}+xz+{{x}^{2}}\And {{y}^{2}}+yz+{{z}^{2}}$is $3ad$ and the constant term is $3{{a}^{2}}+d$ .

Hence, ${{x}^{2}}+xy+{{y}^{2}}\And {{z}^{2}}+xz+{{x}^{2}}\And {{y}^{2}}+yz+{{z}^{2}}$ are in AP with $a'=3{{a}^{2}}+{{d}^{2}}$ and the common difference is $3ad$ .

Now, we know, y = a , x = a-d and z = a + d. So, we can write z – x = 2d.

$\Rightarrow d=\dfrac{z-x}{2}$ .

Substituting the value of a and d in the expression of $a'$ , we get:

$a'=3{{y}^{2}}+\dfrac{{{\left( z-x \right)}^{2}}}{4}$

$\Rightarrow a'=\dfrac{12{{y}^{2}}+{{\left( z-x \right)}^{2}}}{4}$

Also, on substituting the value of a and d in the expression of common difference, we get:

Common difference = $3\times y\times \dfrac{\left( z-x \right)}{2}$

$\Rightarrow Common\,difference=\dfrac{3}{2}\left( z-x \right)y$

Hence, ${{x}^{2}}+xy+{{y}^{2}}\And {{z}^{2}}+xz+{{x}^{2}}\And {{y}^{2}}+yz+{{z}^{2}}$ are in AP with first term $a'=\dfrac{12{{y}^{2}}+{{\left( z-x \right)}^{2}}}{4}$ and common difference = $\dfrac{3}{2}\left( z-x \right)y$ .

Note: One may also try to show that equations ${{x}^{2}}+xy+{{y}^{2}}\And {{z}^{2}}+xz+{{x}^{2}}\And {{y}^{2}}+yz+{{z}^{2}}$ are in AP by using \[y=\dfrac{x+z}{2}\] (by substituting the value of y) or one can also be proved by showing $\left( {{x}^{2}}+xy+{{y}^{2}} \right)-\left( {{z}^{2}}+xz+{{x}^{2}} \right)=\left( {{z}^{2}}+xz+{{x}^{2}} \right)-\left( {{y}^{2}}+yz+{{z}^{2}} \right)$ if $z-y=y-x$ .

We all know that if there are three numbers in AP then they can be written as $\left( a-d \right),a$ and $\left( a+d \right)$. So, let us assume y=a, x=a-d and z=a+d.

We will now put x, y and z in the equation ${{x}^{2}}+xy+{{y}^{2}}$ .

${{\left( a-d \right)}^{2}}+\left( a-d \right)a+{{a}^{2}}$

Upon simplifying we get,

$={{a}^{2}}+{{d}^{2}}-2ad+{{a}^{2}}-ad+{{a}^{2}}$

$=3{{a}^{2}}+{{d}^{2}}-3ad.........\left( i \right)$

Similarly, we will put these values in ${{z}^{2}}+xz+{{x}^{2}}$ .

\[\begin{align}

& {{\left( a+d \right)}^{2}}+\left( a-d \right)\left( a+d \right)+{{\left( a-d \right)}^{2}} \\

& ={{a}^{2}}+{{d}^{2}}+2ad+{{a}^{2}}-{{d}^{2}}+{{a}^{2}}+{{d}^{2}}-2ad \\

& =3{{a}^{2}}+{{d}^{2}}..........\left( ii \right) \\

\end{align}\]

Now, we will put the values of x, y and z in the equation ${{y}^{2}}+yz+{{z}^{2}}$ .

\[\begin{align}

& {{a}^{2}}+a\left( a+d \right)+{{\left( a+d \right)}^{2}} \\

& ={{a}^{2}}+{{a}^{2}}+ad+{{a}^{2}}+{{d}^{2}}+2ad \\

& =3{{a}^{2}}+{{d}^{2}}+3ad..........\left( iii \right) \\

\end{align}\]

Now we will find the difference between equation (i) and equation (ii).

$\left( 3{{a}^{2}}+{{d}^{2}} \right)-\left( 3{{a}^{2}}+{{d}^{2}}-3ad \right)=3ad$

Similarly, we can find the difference between equation (ii) and equation (iii).

$\left( 3{{a}^{2}}+{{d}^{2}}+3ad \right)-\left( 3{{a}^{2}}+{{d}^{2}} \right)=3ad$

We can see from the above equations that the common difference between ${{x}^{2}}+xy+{{y}^{2}}\And {{z}^{2}}+xz+{{x}^{2}}\And {{y}^{2}}+yz+{{z}^{2}}$is $3ad$ and the constant term is $3{{a}^{2}}+d$ .

Hence, ${{x}^{2}}+xy+{{y}^{2}}\And {{z}^{2}}+xz+{{x}^{2}}\And {{y}^{2}}+yz+{{z}^{2}}$ are in AP with $a'=3{{a}^{2}}+{{d}^{2}}$ and the common difference is $3ad$ .

Now, we know, y = a , x = a-d and z = a + d. So, we can write z – x = 2d.

$\Rightarrow d=\dfrac{z-x}{2}$ .

Substituting the value of a and d in the expression of $a'$ , we get:

$a'=3{{y}^{2}}+\dfrac{{{\left( z-x \right)}^{2}}}{4}$

$\Rightarrow a'=\dfrac{12{{y}^{2}}+{{\left( z-x \right)}^{2}}}{4}$

Also, on substituting the value of a and d in the expression of common difference, we get:

Common difference = $3\times y\times \dfrac{\left( z-x \right)}{2}$

$\Rightarrow Common\,difference=\dfrac{3}{2}\left( z-x \right)y$

Hence, ${{x}^{2}}+xy+{{y}^{2}}\And {{z}^{2}}+xz+{{x}^{2}}\And {{y}^{2}}+yz+{{z}^{2}}$ are in AP with first term $a'=\dfrac{12{{y}^{2}}+{{\left( z-x \right)}^{2}}}{4}$ and common difference = $\dfrac{3}{2}\left( z-x \right)y$ .

Note: One may also try to show that equations ${{x}^{2}}+xy+{{y}^{2}}\And {{z}^{2}}+xz+{{x}^{2}}\And {{y}^{2}}+yz+{{z}^{2}}$ are in AP by using \[y=\dfrac{x+z}{2}\] (by substituting the value of y) or one can also be proved by showing $\left( {{x}^{2}}+xy+{{y}^{2}} \right)-\left( {{z}^{2}}+xz+{{x}^{2}} \right)=\left( {{z}^{2}}+xz+{{x}^{2}} \right)-\left( {{y}^{2}}+yz+{{z}^{2}} \right)$ if $z-y=y-x$ .

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