# 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

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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.

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.

Last updated date: 28th Sep 2023

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