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Question: Show that \({{x}^{2}}+xy+{{y}^{2}}\And {{z}^{2}}+xz+{{x}^{2}}\And {{y}^{2}}+yz+{{z}^{2}}\) are in A....

Show that x2+xy+y2&z2+xz+x2&y2+yz+z2{{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.

Explanation

Solution

Hint: We will try to find out the common difference between x2+xy+y2&z2+xz+x2&y2+yz+z2{{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 (a−d),a\left( a-d \right),a and (a+d)\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 x2+xy+y2{{x}^{2}}+xy+{{y}^{2}} .
(a−d)2+(a−d)a+a2{{\left( a-d \right)}^{2}}+\left( a-d \right)a+{{a}^{2}}
Upon simplifying we get,
=a2+d2−2ad+a2−ad+a2={{a}^{2}}+{{d}^{2}}-2ad+{{a}^{2}}-ad+{{a}^{2}}
=3a2+d2−3ad.........(i)=3{{a}^{2}}+{{d}^{2}}-3ad.........\left( i \right)
Similarly, we will put these values in z2+xz+x2{{z}^{2}}+xz+{{x}^{2}} .

& {{\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{aligned}$$ Now, we will put the values of x, y and z in the equation ${{y}^{2}}+yz+{{z}^{2}}$ . $$\begin{aligned} & {{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{aligned}$$ Now we will find the difference between equation (i) and equation (ii). $\left( 3{{a}^{2}}+{{d}^{2}} \right)-\left( 3{{a}^{2}}+{{d}^{2}}-3ad \right)=3ad$ Similarly, we can find the difference between equation (ii) and equation (iii). $\left( 3{{a}^{2}}+{{d}^{2}}+3ad \right)-\left( 3{{a}^{2}}+{{d}^{2}} \right)=3ad$ We can see from the above equations that the common difference between ${{x}^{2}}+xy+{{y}^{2}}\And {{z}^{2}}+xz+{{x}^{2}}\And {{y}^{2}}+yz+{{z}^{2}}$is $3ad$ and the constant term is $3{{a}^{2}}+{d}^{2}$ . Hence, ${{x}^{2}}+xy+{{y}^{2}}\And {{z}^{2}}+xz+{{x}^{2}}\And {{y}^{2}}+yz+{{z}^{2}}$ are in AP with $a'=3{{a}^{2}}+{{d}^{2}}$ and the common difference is $3ad$ . Now, we know, y = a , x = a-d and z = a + d. So, we can write z – x = 2d. $\Rightarrow d=\dfrac{z-x}{2}$ . Substituting the value of a and d in the expression of $a'$ , we get: $a'=3{{y}^{2}}+\dfrac{{{\left( z-x \right)}^{2}}}{4}$ $\Rightarrow a'=\dfrac{12{{y}^{2}}+{{\left( z-x \right)}^{2}}}{4}$ Also, on substituting the value of a and d in the expression of common difference, we get: Common difference = $3\times y\times \dfrac{\left( z-x \right)}{2}$ $\Rightarrow Common\,difference=\dfrac{3}{2}\left( z-x \right)y$ Hence, ${{x}^{2}}+xy+{{y}^{2}}\And {{z}^{2}}+xz+{{x}^{2}}\And {{y}^{2}}+yz+{{z}^{2}}$ are in AP with first term $a'=\dfrac{12{{y}^{2}}+{{\left( z-x \right)}^{2}}}{4}$ and common difference = $\dfrac{3}{2}\left( z-x \right)y$ . Note: One may also try to show that equations ${{x}^{2}}+xy+{{y}^{2}}\And {{z}^{2}}+xz+{{x}^{2}}\And {{y}^{2}}+yz+{{z}^{2}}$ are in AP by using $$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 $z-y=y-x$ .