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# Show that ${{x}^{2}}+xy+{{y}^{2}}\And {{z}^{2}}+xz+{{x}^{2}}\And {{y}^{2}}+yz+{{z}^{2}}$ are in A.P. if x, y, z are in A.P.

Answer Verified
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 $z-y=y-x$ .