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

Hint: 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
\[\begin{align}
  & \Rightarrow {{\left( by \right)}^{2}}=\left( ax \right)\left( cz \right) \\
 & \Rightarrow {{y}^{2}}=\dfrac{acxz}{{{b}^{2}}}..........equation(ii) \\
\end{align}\]
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
\[\begin{align}
  & \Rightarrow \dfrac{2}{b}=\dfrac{1}{a}+\dfrac{1}{c} \\
 & \Rightarrow \dfrac{2}{b}=\dfrac{a+c}{ac}.........equation(iii) \\
\end{align}\]
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
\[\begin{align}
  & \Rightarrow {{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}} \\
 & \Rightarrow {{a}^{2}}+{{b}^{2}}={{\left( a+b \right)}^{2}}-2ab \\
\end{align}\]
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
\[\begin{align}
  & \Rightarrow {{\left( 2y \right)}^{2}}={{\left( x+z \right)}^{2}} \\
 & \Rightarrow {{\left( x+z \right)}^{2}}=4{{y}^{2}} \\
\end{align}\]
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
\[\begin{align}
  & \Rightarrow LHS=\dfrac{\left( \dfrac{4acxz}{{{b}^{2}}} \right)-2xz}{xz} \\
 & \Rightarrow LHS=\dfrac{4ac}{{{b}^{2}}}-2........equation(v) \\
\end{align}\]
Now, let us take the equation (iii) and by squaring on both sides we get
\[\begin{align}
  & \Rightarrow {{\left( \dfrac{2}{b} \right)}^{2}}={{\left( \dfrac{a+c}{ac} \right)}^{2}} \\
 & \Rightarrow \dfrac{4}{{{b}^{2}}}=\dfrac{{{\left( a+c \right)}^{2}}}{{{\left( ac \right)}^{2}}} \\
\end{align}\]
By substituting the value of \[\dfrac{4}{{{b}^{2}}}\] in equation (v) we get
\[\begin{align}
  & \Rightarrow LHS=ac\left( \dfrac{{{\left( a+c \right)}^{2}}}{{{\left( ac \right)}^{2}}} \right)-2 \\
 & \Rightarrow LHS=\dfrac{{{\left( a+c \right)}^{2}}-2ac}{ac} \\
\end{align}\]
By expanding the square of sum of two numbers in above formula we get
\[\begin{align}
  & \Rightarrow LHS=\dfrac{{{a}^{2}}+{{c}^{2}}+2ac-2ac}{ac} \\
 & \Rightarrow LHS=\dfrac{{{a}^{2}}}{ac}+\dfrac{{{c}^{2}}}{ac} \\
 & \Rightarrow LHS=\dfrac{a}{c}+\dfrac{c}{a} \\
\end{align}\]
Now let us take the RHS of given result that is
\[\begin{align}
  & \Rightarrow RHS=\dfrac{a}{c}+\dfrac{c}{a} \\
 & \Rightarrow RHS=LHS \\
\end{align}\]
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
\[\begin{align}
  & \Rightarrow RHS=\dfrac{a}{c}+\dfrac{c}{a} \\
 & \Rightarrow RHS=\dfrac{{{a}^{2}}+{{c}^{2}}}{ac} \\
\end{align}\]
We know that the formula of square of two numbers as
\[\begin{align}
  & \Rightarrow {{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}} \\
 & \Rightarrow {{a}^{2}}+{{b}^{2}}={{\left( a+b \right)}^{2}}-2ab \\
\end{align}\]
By using this result in the above equation we get
\[\begin{align}
  & \Rightarrow RHS=\dfrac{{{\left( a+c \right)}^{2}}-2ac}{ac} \\
 & \Rightarrow RHS=\dfrac{{{\left( a+c \right)}^{2}}}{ac}-2 \\
\end{align}\]
Now, let us take the equation (iii) and by squaring on both sides we get
\[\begin{align}
  & \Rightarrow {{\left( \dfrac{2}{b} \right)}^{2}}={{\left( \dfrac{a+c}{ac} \right)}^{2}} \\
 & \Rightarrow \dfrac{4}{{{b}^{2}}}=\dfrac{{{\left( a+c \right)}^{2}}}{{{\left( ac \right)}^{2}}} \\
 & \Rightarrow \dfrac{{{\left( a+c \right)}^{2}}}{\left( ac \right)}=\dfrac{4ac}{{{b}^{2}}} \\
\end{align}\]
By using this result in the above equation we get
\[\begin{align}
  & \Rightarrow RHS=\dfrac{4ac}{{{b}^{2}}}-2 \\
 & \Rightarrow RHS=LHS \\
\end{align}\]
Hence the required result has been proved.