Question

If $x,y,z$ are in A.P, $ax,by,cz$ are in G.P and $a,b,c$ are in H.P then prove that
$\dfrac{x}{z}+\dfrac{z}{x}=\dfrac{a}{c}+\dfrac{c}{a}$

Answer 1

We solve this problem by using the conditions of A.P, G.P and H.P.
If $a,b,c$ are in A.P then $2b=a+c$
If $a,b,c$ are in G.P then ${{b}^{2}}=ac$
If $a,b,c$ are in H.P then $\dfrac{2}{b}=\dfrac{1}{a}+\dfrac{1}{b}$
By using the above three conditions we solve for the required result.

Complete step by step answer:
We are given that $x,y,z$ are in A.P
We know that the condition of A.P as
If $a,b,c$ are in A.P then $2b=a+c$
By using the above condition to given sequence we get
$\Rightarrow 2y=x+z....equation(i)$
We are given that $ax,by,cz$ are in G.P
We know that the condition of G.P as
If $a,b,c$ are in G.P then ${{b}^{2}}=ac$
By using the above condition to given sequence we get

& \Rightarrow {{\left( by \right)}^{2}}=\left( ax \right)\left( cz \right) \\\ & \Rightarrow {{y}^{2}}=\dfrac{acxz}{{{b}^{2}}}..........equation(ii) \\\ \end{aligned}$$ We are given that $$a,b,c$$ are in H.P We know that the condition of H.P as If $$a,b,c$$ are in H.P then $$\dfrac{2}{b}=\dfrac{1}{a}+\dfrac{1}{b}$$ By using the above condition to given sequence we get $$\begin{aligned} & \Rightarrow \dfrac{2}{b}=\dfrac{1}{a}+\dfrac{1}{c} \\\ & \Rightarrow \dfrac{2}{b}=\dfrac{a+c}{ac}.........equation(iii) \\\ \end{aligned}$$ We are asked to prove that $$\dfrac{x}{z}+\dfrac{z}{x}=\dfrac{a}{c}+\dfrac{c}{a}$$ Now, let us take the LHS as $$\Rightarrow LHS=\dfrac{x}{z}+\dfrac{z}{x}$$ By doing LCM and adding the fractions we get $$\Rightarrow LHS=\dfrac{{{x}^{2}}+{{z}^{2}}}{xz}$$ We know that the formula of square of two numbers as $$\begin{aligned} & \Rightarrow {{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}} \\\ & \Rightarrow {{a}^{2}}+{{b}^{2}}={{\left( a+b \right)}^{2}}-2ab \\\ \end{aligned}$$ By using this result in the above equation we get $$\Rightarrow LHS=\dfrac{{{\left( x+z \right)}^{2}}-2xz}{xz}........equation(iv)$$ Now, let us take the equation (i) and by squaring on both sides we get $$\begin{aligned} & \Rightarrow {{\left( 2y \right)}^{2}}={{\left( x+z \right)}^{2}} \\\ & \Rightarrow {{\left( x+z \right)}^{2}}=4{{y}^{2}} \\\ \end{aligned}$$ We know that from the equation (ii) that is $$\Rightarrow {{y}^{2}}=\dfrac{acxz}{{{b}^{2}}}$$ By substituting this value in above equation we get $$\Rightarrow {{\left( x+z \right)}^{2}}=\dfrac{4acxz}{{{b}^{2}}}$$ Now, by substituting the value of $${{\left( x+z \right)}^{2}}$$ in equation (iv) we get $$\begin{aligned} & \Rightarrow LHS=\dfrac{\left( \dfrac{4acxz}{{{b}^{2}}} \right)-2xz}{xz} \\\ & \Rightarrow LHS=\dfrac{4ac}{{{b}^{2}}}-2........equation(v) \\\ \end{aligned}$$ Now, let us take the equation (iii) and by squaring on both sides we get $$\begin{aligned} & \Rightarrow {{\left( \dfrac{2}{b} \right)}^{2}}={{\left( \dfrac{a+c}{ac} \right)}^{2}} \\\ & \Rightarrow \dfrac{4}{{{b}^{2}}}=\dfrac{{{\left( a+c \right)}^{2}}}{{{\left( ac \right)}^{2}}} \\\ \end{aligned}$$ By substituting the value of $$\dfrac{4}{{{b}^{2}}}$$ in equation (v) we get $$\begin{aligned} & \Rightarrow LHS=ac\left( \dfrac{{{\left( a+c \right)}^{2}}}{{{\left( ac \right)}^{2}}} \right)-2 \\\ & \Rightarrow LHS=\dfrac{{{\left( a+c \right)}^{2}}-2ac}{ac} \\\ \end{aligned}$$ By expanding the square of sum of two numbers in above formula we get $$\begin{aligned} & \Rightarrow LHS=\dfrac{{{a}^{2}}+{{c}^{2}}+2ac-2ac}{ac} \\\ & \Rightarrow LHS=\dfrac{{{a}^{2}}}{ac}+\dfrac{{{c}^{2}}}{ac} \\\ & \Rightarrow LHS=\dfrac{a}{c}+\dfrac{c}{a} \\\ \end{aligned}$$ Now let us take the RHS of given result that is $$\begin{aligned} & \Rightarrow RHS=\dfrac{a}{c}+\dfrac{c}{a} \\\ & \Rightarrow RHS=LHS \\\ \end{aligned}$$ Hence the required result has been proved. **Note:** We can solve this problem in other way. Here, we get the equation as $$\Rightarrow LHS=\dfrac{4ac}{{{b}^{2}}}-2$$ Let us go for RHS now that is $$\begin{aligned} & \Rightarrow RHS=\dfrac{a}{c}+\dfrac{c}{a} \\\ & \Rightarrow RHS=\dfrac{{{a}^{2}}+{{c}^{2}}}{ac} \\\ \end{aligned}$$ We know that the formula of square of two numbers as $$\begin{aligned} & \Rightarrow {{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}} \\\ & \Rightarrow {{a}^{2}}+{{b}^{2}}={{\left( a+b \right)}^{2}}-2ab \\\ \end{aligned}$$ By using this result in the above equation we get $$\begin{aligned} & \Rightarrow RHS=\dfrac{{{\left( a+c \right)}^{2}}-2ac}{ac} \\\ & \Rightarrow RHS=\dfrac{{{\left( a+c \right)}^{2}}}{ac}-2 \\\ \end{aligned}$$ Now, let us take the equation (iii) and by squaring on both sides we get $$\begin{aligned} & \Rightarrow {{\left( \dfrac{2}{b} \right)}^{2}}={{\left( \dfrac{a+c}{ac} \right)}^{2}} \\\ & \Rightarrow \dfrac{4}{{{b}^{2}}}=\dfrac{{{\left( a+c \right)}^{2}}}{{{\left( ac \right)}^{2}}} \\\ & \Rightarrow \dfrac{{{\left( a+c \right)}^{2}}}{\left( ac \right)}=\dfrac{4ac}{{{b}^{2}}} \\\ \end{aligned}$$ By using this result in the above equation we get $$\begin{aligned} & \Rightarrow RHS=\dfrac{4ac}{{{b}^{2}}}-2 \\\ & \Rightarrow RHS=LHS \\\ \end{aligned}$$ Hence the required result has been proved.