
If three unequal non-zero real numbers $a,b,c$ are in G.P. and $b-c,c-a,a-b$ are in H.P., then the value of $a+b+c$ is independent of
A. $a$
B. $b$
C. $c$
D. None of these
Answer
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Hint: In this question, we are to find the result of the expression $a+b+c$, which is formed by applying appropriate operations on the terms of a G.P. series $a,b,c$ and the terms of a H.P. series $b-c,c-a,a-b$. By using the means, the required expression is evaluated.
Formula used :
The geometric mean of a series with the terms $a,b,c$ is $G.M=b=\sqrt{ac}$
The harmonic mean of a series with the terms $a,b,c$ is $H.M=b=\frac{2ac}{a+c}$
Complete Step-by-Step Solution:
Given that, the terms $a,b,c$ are in G.P. and $b-c,c-a,a-b$ are in H.P.
So, we can find their means as follows:
The geometric mean of the given terms which are in G.P. is
$\begin{align}
& b=\sqrt{ac} \\
& \Rightarrow {{b}^{2}}=ac\text{ }...(1) \\
\end{align}$
The harmonic mean of the given terms which are in H.P. is
$(c-a)=\frac{2(b-c)(a-b)}{(b-c)+(a-b)}$
On simplifying the obtained H.M., we get
$\begin{align}
& (c-a)=\frac{2(b-c)(a-b)}{(b-c)+(a-b)} \\
& \Rightarrow (c-a)=\frac{2(ab-{{b}^{2}}-ac+bc)}{a-c} \\
& \Rightarrow (a-c)(c-a)=2(ab-{{b}^{2}}-ac+bc) \\
\end{align}$
Substituting (1) in the above equation, we get
$\begin{align}
& \Rightarrow -(a-c)(a-c)=2(ab-{{b}^{2}}-{{b}^{2}}+bc) \\
& \Rightarrow -(a-c)(a-c)=2(ab-2{{b}^{2}}+bc) \\
\end{align}$
We can write the L.H.S as
${{(a-c)}^{2}}=(a-c)(a-c)=\left\{ {{\left( \sqrt{a}-\sqrt{c} \right)}^{2}}{{\left( \sqrt{a}+\sqrt{c} \right)}^{2}} \right\}$
So, we get
\[\begin{align}
& -{{\left( \sqrt{a}-\sqrt{c} \right)}^{2}}{{\left( \sqrt{a}+\sqrt{c} \right)}^{2}}=2b(a-2b+c) \\
& \Rightarrow -{{\left( \sqrt{a}-\sqrt{c} \right)}^{2}}{{\left( \sqrt{a}+\sqrt{c} \right)}^{2}}=2b(a-2\sqrt{ac}+c) \\
& \because b=\sqrt{ac};(R.H.S) \\
& \Rightarrow -{{\left( \sqrt{a}-\sqrt{c} \right)}^{2}}{{\left( \sqrt{a}+\sqrt{c} \right)}^{2}}=2b\left( {{\left( \sqrt{a} \right)}^{2}}-2\sqrt{a}\cdot \sqrt{c}+{{\left( \sqrt{c} \right)}^{2}} \right) \\
& \Rightarrow -{{\left( \sqrt{a}-\sqrt{c} \right)}^{2}}{{\left( \sqrt{a}+\sqrt{c} \right)}^{2}}=2b{{\left( \sqrt{a}-\sqrt{c} \right)}^{2}} \\
\end{align}\]
On simplifying, we get
$\begin{align}
& -{{\left( \sqrt{a}+\sqrt{c} \right)}^{2}}=2b \\
& \Rightarrow -(a+2\sqrt{ac}+c)=2b \\
& \Rightarrow -(a+2b+c)=2b \\
& \Rightarrow a+b+c=-2b-b=-3b \\
& \therefore a+b+c=-3b=-3\sqrt{ac} \\
\end{align}$
Thus, we can write that the expression is dependent on the values of a, b, c. So, the correct option is “D - none of the above”.
Hence, the option D is correct
Note:
In this problem, we need to remember that, the given terms are under progressions. So, we can relate them by finding their means. Applying this method or strategy to the above terms we get the required expression and its result. So, we can conclude that it depends on all the terms from the above.
Formula used :
The geometric mean of a series with the terms $a,b,c$ is $G.M=b=\sqrt{ac}$
The harmonic mean of a series with the terms $a,b,c$ is $H.M=b=\frac{2ac}{a+c}$
Complete Step-by-Step Solution:
Given that, the terms $a,b,c$ are in G.P. and $b-c,c-a,a-b$ are in H.P.
So, we can find their means as follows:
The geometric mean of the given terms which are in G.P. is
$\begin{align}
& b=\sqrt{ac} \\
& \Rightarrow {{b}^{2}}=ac\text{ }...(1) \\
\end{align}$
The harmonic mean of the given terms which are in H.P. is
$(c-a)=\frac{2(b-c)(a-b)}{(b-c)+(a-b)}$
On simplifying the obtained H.M., we get
$\begin{align}
& (c-a)=\frac{2(b-c)(a-b)}{(b-c)+(a-b)} \\
& \Rightarrow (c-a)=\frac{2(ab-{{b}^{2}}-ac+bc)}{a-c} \\
& \Rightarrow (a-c)(c-a)=2(ab-{{b}^{2}}-ac+bc) \\
\end{align}$
Substituting (1) in the above equation, we get
$\begin{align}
& \Rightarrow -(a-c)(a-c)=2(ab-{{b}^{2}}-{{b}^{2}}+bc) \\
& \Rightarrow -(a-c)(a-c)=2(ab-2{{b}^{2}}+bc) \\
\end{align}$
We can write the L.H.S as
${{(a-c)}^{2}}=(a-c)(a-c)=\left\{ {{\left( \sqrt{a}-\sqrt{c} \right)}^{2}}{{\left( \sqrt{a}+\sqrt{c} \right)}^{2}} \right\}$
So, we get
\[\begin{align}
& -{{\left( \sqrt{a}-\sqrt{c} \right)}^{2}}{{\left( \sqrt{a}+\sqrt{c} \right)}^{2}}=2b(a-2b+c) \\
& \Rightarrow -{{\left( \sqrt{a}-\sqrt{c} \right)}^{2}}{{\left( \sqrt{a}+\sqrt{c} \right)}^{2}}=2b(a-2\sqrt{ac}+c) \\
& \because b=\sqrt{ac};(R.H.S) \\
& \Rightarrow -{{\left( \sqrt{a}-\sqrt{c} \right)}^{2}}{{\left( \sqrt{a}+\sqrt{c} \right)}^{2}}=2b\left( {{\left( \sqrt{a} \right)}^{2}}-2\sqrt{a}\cdot \sqrt{c}+{{\left( \sqrt{c} \right)}^{2}} \right) \\
& \Rightarrow -{{\left( \sqrt{a}-\sqrt{c} \right)}^{2}}{{\left( \sqrt{a}+\sqrt{c} \right)}^{2}}=2b{{\left( \sqrt{a}-\sqrt{c} \right)}^{2}} \\
\end{align}\]
On simplifying, we get
$\begin{align}
& -{{\left( \sqrt{a}+\sqrt{c} \right)}^{2}}=2b \\
& \Rightarrow -(a+2\sqrt{ac}+c)=2b \\
& \Rightarrow -(a+2b+c)=2b \\
& \Rightarrow a+b+c=-2b-b=-3b \\
& \therefore a+b+c=-3b=-3\sqrt{ac} \\
\end{align}$
Thus, we can write that the expression is dependent on the values of a, b, c. So, the correct option is “D - none of the above”.
Hence, the option D is correct
Note:
In this problem, we need to remember that, the given terms are under progressions. So, we can relate them by finding their means. Applying this method or strategy to the above terms we get the required expression and its result. So, we can conclude that it depends on all the terms from the above.
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