Question

If a, b, c in the G.P. and ${{a}^{x}}={{b}^{y}}={{c}^{z}}$ then prove that x, y, z are in H.P

Answer 1

Hint: use the basic definition of G.P. i.e. if three terms (a, b, c) are in G.P., then, relation between (a, b, c) are ${{b}^{2}}=ac.$

We are given that
${{a}^{x}}={{b}^{y}}={{c}^{z}}...........\left( 1 \right)$
Another information given in the question is that a, b, c are G.P. and hence, we know that, if three terms (a, b, c) in G.P., then can write relation between then as;
$\begin{align}
  & {{b}^{2}}=ac \\
 & Or \\
 & b=\sqrt{ac}........................\left( 2 \right) \\
\end{align}$
Now we can substitute value of ‘b’ from equation (2) to equation (1), we get;
$\begin{align}
  & {{a}^{x}}={{\left( \sqrt{ac} \right)}^{y}}={{c}^{z}} \\
 & or \\
 & {{a}^{x}}={{\left( ac \right)}^{\dfrac{y}{2}}}={{c}^{z}}...............\left( 3 \right) \\
\end{align}$
Let us solve the first two terms and second terms individually to get a relation among x, y, z.
Now, from first two terms of equation (3), we get;
${{a}^{x}}={{\left( ac \right)}^{\dfrac{y}{2}}}...........\left( 4 \right)$
As we know property of surds that
${{\left( mn \right)}^{r}}={{m}^{r}}{{n}^{r}}$
Hence, we can simplify equation (4), as
${{a}^{x}}={{a}^{\dfrac{y}{2}}}{{c}^{\dfrac{y}{2}}}$
Transferring ${{a}^{\dfrac{y}{2}}}$ to other sides, we get;
$\dfrac{{{a}^{x}}}{{{a}^{\dfrac{y}{2}}}}={{c}^{\dfrac{y}{2}}}................\left( 5 \right)$
Now, using property of surds as
$\dfrac{{{m}^{r}}}{{{m}^{n}}}={{m}^{r-n}}$
Now, equation (5), becomes
${{a}^{x-\dfrac{y}{2}}}={{c}^{\dfrac{y}{2}}}.................\left( 6 \right)$
Now, taking last two terms of equation (3), we get;
${{\left( ac \right)}^{\dfrac{y}{2}}}={{c}^{z}}$
Now we can simplify the above relation using ${{\left( mn \right)}^{r}}={{m}^{r}}{{n}^{r}}$. Hence, above equation can be written as;
$\begin{align}
  & {{\left( ac \right)}^{\dfrac{y}{2}}}={{c}^{z}} \\
 & {{a}^{\dfrac{y}{2}}}{{c}^{\dfrac{y}{2}}}={{c}^{z}} \\
\end{align}$
Transferring ${{c}^{\dfrac{y}{2}}}$ to other sides, we get;
${{a}^{\dfrac{y}{2}}}=\dfrac{{{c}^{z}}}{{{c}^{\dfrac{y}{2}}}}$
Now, using property, $\dfrac{{{m}^{r}}}{{{m}^{n}}}={{m}^{r-n}}$, we can rewrite the given equation as;
${{a}^{\dfrac{y}{2}}}={{c}^{z-\dfrac{y}{2}}}..............\left( 7 \right)$
Now we know the property of surds as,
If \[{{a}^{m}}={{b}^{n}}\], we can transfer power to other side as
\[a={{\left( {{b}^{n}} \right)}^{\dfrac{1}{m}}}\text{ or }a={{b}^{\dfrac{n}{m}}}..............\left( 8 \right)\]
Using the above property of equation (8), with the equation (7), we get equation (7) as
$\begin{align}
  & a={{c}^{\left( z-\dfrac{y}{2} \right)\dfrac{1}{\left( \dfrac{y}{2} \right)}}} \\
 & or \\
 & a={{c}^{\left( z-\dfrac{y}{2} \right)\dfrac{2}{y}}}..............\left( 9 \right) \\
\end{align}$
Now putting value of ‘a’ to equation (6) we get;
${{c}^{\left( \left( z-\dfrac{y}{2} \right)\dfrac{2}{y} \right)\left( x-\dfrac{y}{2} \right)}}={{c}^{\dfrac{y}{2}}}..................\left( 10 \right)$
Using the property of surds that if ${{a}^{m}}={{a}^{n}}$ then power should also be equal i.e. m=n.
Therefore, we can write from equation (10),
$\left( z-\dfrac{y}{2} \right)\dfrac{2}{y}\left( x-\dfrac{y}{2} \right)=\dfrac{y}{2}$
On simplifying the above relation, we get
 $\begin{align}
  & \left( \dfrac{2z-y}{2} \right)\left( \dfrac{2}{y} \right)\left( \dfrac{2x-y}{2} \right)=\dfrac{y}{2} \\
 & \left( 2z-y \right)\left( 2x-y \right)={{y}^{2}} \\
\end{align}$
Multiplying (2z – y) and (2x – y), we get
$\begin{align}
  & 4xz-2yz-2xy+{{y}^{2}}={{y}^{2}} \\
 & 4xz-2yz-2xy=0 \\
\end{align}$
Dividing the whole equation by 2, we get
2xz – yz – xy = 0
Or
xy + yz = 2xz
Dividing, the whole equation by xyz to both sides, we get,
$\begin{align}
  & \dfrac{xy}{xyz}+\dfrac{yz}{xyz}=\dfrac{2xz}{xyz} \\
 & \dfrac{1}{z}+\dfrac{1}{x}=\dfrac{2}{y} \\
\end{align}$
As we know that if three numbers x, y, z are in HP, then $\dfrac{1}{x}+\dfrac{1}{z}=\dfrac{2}{y}$as written in above equation.
Hence, it is proved that x, y, z are in H.P.

Note: One can go wrong while using the property if \[{{a}^{m}}={{c}^{n}}\] then \[a={{c}^{\dfrac{n}{m}}}\].
One can go wrong while transferring m to other side as
If \[{{a}^{m}}={{c}^{n}}\]then \[a={{c}^{\dfrac{n}{m}}}\text{ or }a={{c}^{mn}}\] which are wrong. Hence, be careful while using the above property of surds.
Another approach for this question would be that can take log to equation as,
${{a}^{x}}={{b}^{y}}={{c}^{z}}$
Taking log and using property as
$\log {{m}^{n}}=n\log m$
$\begin{align}
  & \log {{a}^{x}}=\log {{b}^{y}}=\log {{c}^{z}} \\
 & x\log a=y\log b=z\log c \\
\end{align}$
We know, ${{b}^{2}}=ac$
Taking log to both sides, we get;
$\begin{align}
  & \log {{b}^{2}}=\log ac \\
 & 2\log b=\log a+\log c \\
\end{align}$
As, we know log ab = log a + log c
Now, using the two equations
$x\log a=y\log b=z\log c\text{ and }2\log b=\log a+\log c,$ find relation between x, y and z.