If a,b,c are in A.P. and geometric means of $ac$ and $ab$ , $ab$ and $bc$ , $ca$ and $cb$ are $d,e,f$ respectively then ${{d}^{2}},{{e}^{2}},{{f}^{2}}$ are in:
(1) A.P.
(2) G.P.
(3) H.P.
(4) A.G.P.
Answer
526.2k+ views
Hint: Here in this question, we have been given the following information “$a,b,c$ are in A.P. and geometric means of $ac$ and $ab$ , $ab$ and $bc$ , $ca$ and $cb$ are $d,e,f$ respectively” and asked to find the name of the progression ${{d}^{2}},{{e}^{2}},{{f}^{2}}$. Consecutive terms in A.P. have a common difference.
Complete step by step solution:
Now considering from the question we have been given that “$a,b,c$ are in A.P. and geometric means of $ac$ and $ab$ , $ab$ and $bc$ , $ca$ and $cb$ are $d,e,f$ respectively”.
From the basic concepts, we know that the consecutive terms in A.P. have a common difference.
Hence we can say that $b-a=c-b\Rightarrow 2b=a+c$ .
From the basic concepts, we know that the geometric mean of any two numbers $p,q$ is given as $\sqrt{pq}$ .
Hence we can say that $\sqrt{ab\times ac}=d$, $\sqrt{ab\times bc}=e$and $\sqrt{cb\times ca}=f$ .
Hence we can say that ${{d}^{2}}={{a}^{2}}\left( bc \right)$ , ${{e}^{2}}={{b}^{2}}\left( ac \right)$ and ${{f}^{2}}={{c}^{2}}\left( ab \right)$ .
Let us simplify it further then we will have
$\begin{align}
& {{d}^{2}}+{{f}^{2}}={{a}^{2}}\left( bc \right)+{{c}^{2}}\left( ab \right) \\
& \Rightarrow {{d}^{2}}+{{f}^{2}}=\left( abc \right)\left( a+c \right) \\
\end{align}$ .
Let us use $b=\dfrac{a+c}{2}$ then we will have $\Rightarrow {{d}^{2}}+{{f}^{2}}=\left( \dfrac{ac}{2} \right){{\left( a+c \right)}^{2}}$ .
Now let us use $b=\dfrac{a+c}{2}$ then we will have $\Rightarrow {{e}^{2}}={{\left( \dfrac{a+c}{2} \right)}^{2}}\left( ac \right)$.
Hence we can say that $2{{e}^{2}}={{d}^{2}}+{{f}^{2}}$ .
Therefore we can conclude that if $a,b,c$ are in A.P. and geometric means of $ac$ and $ab$ , $ab$ and $bc$ , $ca$ and $cb$ are $d,e,f$ respectively then ${{d}^{2}},{{e}^{2}},{{f}^{2}}$ are in A.P.
Hence we will mark the option “1” as correct.
Note: While answering questions of this type, we should be sure of the concepts that we are going to use in between the steps. Very few mistakes are possible in questions of this type. This is directly based on the concept of progressions without involving much calculation.
Complete step by step solution:
Now considering from the question we have been given that “$a,b,c$ are in A.P. and geometric means of $ac$ and $ab$ , $ab$ and $bc$ , $ca$ and $cb$ are $d,e,f$ respectively”.
From the basic concepts, we know that the consecutive terms in A.P. have a common difference.
Hence we can say that $b-a=c-b\Rightarrow 2b=a+c$ .
From the basic concepts, we know that the geometric mean of any two numbers $p,q$ is given as $\sqrt{pq}$ .
Hence we can say that $\sqrt{ab\times ac}=d$, $\sqrt{ab\times bc}=e$and $\sqrt{cb\times ca}=f$ .
Hence we can say that ${{d}^{2}}={{a}^{2}}\left( bc \right)$ , ${{e}^{2}}={{b}^{2}}\left( ac \right)$ and ${{f}^{2}}={{c}^{2}}\left( ab \right)$ .
Let us simplify it further then we will have
$\begin{align}
& {{d}^{2}}+{{f}^{2}}={{a}^{2}}\left( bc \right)+{{c}^{2}}\left( ab \right) \\
& \Rightarrow {{d}^{2}}+{{f}^{2}}=\left( abc \right)\left( a+c \right) \\
\end{align}$ .
Let us use $b=\dfrac{a+c}{2}$ then we will have $\Rightarrow {{d}^{2}}+{{f}^{2}}=\left( \dfrac{ac}{2} \right){{\left( a+c \right)}^{2}}$ .
Now let us use $b=\dfrac{a+c}{2}$ then we will have $\Rightarrow {{e}^{2}}={{\left( \dfrac{a+c}{2} \right)}^{2}}\left( ac \right)$.
Hence we can say that $2{{e}^{2}}={{d}^{2}}+{{f}^{2}}$ .
Therefore we can conclude that if $a,b,c$ are in A.P. and geometric means of $ac$ and $ab$ , $ab$ and $bc$ , $ca$ and $cb$ are $d,e,f$ respectively then ${{d}^{2}},{{e}^{2}},{{f}^{2}}$ are in A.P.
Hence we will mark the option “1” as correct.
Note: While answering questions of this type, we should be sure of the concepts that we are going to use in between the steps. Very few mistakes are possible in questions of this type. This is directly based on the concept of progressions without involving much calculation.
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