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Question

If $4{a}^{2}+9{b}^{2}+16{c}^{2}=2\left(3ab+6bc+4ca\right)$, where $a,b$ and $c$ are non - zero numbers, then $a,b$ and $c$ are in

A

$AP$

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B

$GP$

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C

$HP$

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D

None of these

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Solution

The correct option is C $HP$Explanation for the correct option:Given $4{a}^{2}+9{b}^{2}+16{c}^{2}=2\left(3ab+6bc+4ca\right)$$⇒$ $\begin{array}{rcl}2\left(4{a}^{2}+9{b}^{2}+16{c}^{2}\right)& =& 2×2\left(3ab+6bc+4ca\right)\end{array}$$⇒$$\begin{array}{rcl}4{a}^{2}+4{a}^{2}+9{b}^{2}+9{b}^{2}+16{c}^{2}+16{c}^{2}-12ab-24bc-16ca& =& 0\end{array}$$⇒$ $\begin{array}{rcl}{\left(2a-3b\right)}^{2}+{\left(3b-4c\right)}^{2}+{\left(4c-2a\right)}^{2}& =& 0\end{array}$Now we have ,$\begin{array}{rcl}2a-3b& =& 0,\\ 3b-4c& =& 0,\\ 4c-2a& =& 0\end{array}$$⇒$$\begin{array}{rcl}\left(2a=3b=4c\right)& =& x\end{array}$ $\begin{array}{rcl}a& =& x/2,\\ b& =& x/3,\\ c& =& x/4\end{array}$ $\frac{1}{c}+\frac{1}{a}=\frac{6}{x}$$⇒$$\frac{1}{c}+\frac{1}{a}=\frac{2}{b}$So $a,b,c$ are in HP.Hence option C is the answer.

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