Questions & Answers

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 Verified Verified
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;
  & {{b}^{2}}=ac \\
 & Or \\
 & b=\sqrt{ac}........................\left( 2 \right) \\
Now we can substitute value of ‘b’ from equation (2) to equation (1), we get;
  & {{a}^{x}}={{\left( \sqrt{ac} \right)}^{y}}={{c}^{z}} \\
 & or \\
 & {{a}^{x}}={{\left( ac \right)}^{\dfrac{y}{2}}}={{c}^{z}}...............\left( 3 \right) \\
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
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
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;
  & {{\left( ac \right)}^{\dfrac{y}{2}}}={{c}^{z}} \\
 & {{a}^{\dfrac{y}{2}}}{{c}^{\dfrac{y}{2}}}={{c}^{z}} \\
Transferring ${{c}^{\dfrac{y}{2}}}$ to other sides, we get;
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
  & 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) \\
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
  & \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}} \\
Multiplying (2z – y) and (2x – y), we get
  & 4xz-2yz-2xy+{{y}^{2}}={{y}^{2}} \\
 & 4xz-2yz-2xy=0 \\
Dividing the whole equation by 2, we get
2xz – yz – xy = 0
xy + yz = 2xz
Dividing, the whole equation by xyz to both sides, we get,
  & \dfrac{xy}{xyz}+\dfrac{yz}{xyz}=\dfrac{2xz}{xyz} \\
 & \dfrac{1}{z}+\dfrac{1}{x}=\dfrac{2}{y} \\
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,
Taking log and using property as
$\log {{m}^{n}}=n\log m$
  & \log {{a}^{x}}=\log {{b}^{y}}=\log {{c}^{z}} \\
 & x\log a=y\log b=z\log c \\
We know, ${{b}^{2}}=ac$
Taking log to both sides, we get;
  & \log {{b}^{2}}=\log ac \\
 & 2\log b=\log a+\log c \\
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.
Bookmark added to your notes.
View Notes