Question

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

• A. $AP$
• B. $GP$
• C. $HP$
• D. None of these

Answer 1

The correct option is C

$HP$

Explanation for the correct option:

Given $4a^2+9b^2+16c^2=2(3ab+6bc+4ca)$

$\Rightarrow$ $\begin{array}{rcl}2(4a^2+9b^2+16c^2)& =& 2\times 2(3ab+6bc+4ca)\end{array}$

$\Rightarrow$$\begin{array}{rcl}4a^2+4a^2+9b^2+9b^2+16c^2+16c^2-12ab-24bc-16ca& =& 0\end{array}$

$\Rightarrow$ $\begin{array}{rcl}{(2a-3b)}^2+{(3b-4c)}^2+{(4c-2a)}^2& =& 0\end{array}$

Now we have ,

$\begin{array}{rcl}2a-3b& =& 0,\\ 3b-4c& =& 0,\\ 4c-2a& =& 0\end{array}$

$\Rightarrow$$\begin{array}{rcl}\left(2a=3b=4c\right)& =& x\end{array}$

$\begin{array}{rcl}a& =& x/2,\\ b& =& x/3,\\ c& =& x/4\end{array}$

$\frac{1}{c}+\frac{1}{a}=\frac{6}{x}$

$\Rightarrow$$\frac{1}{c}+\frac{1}{a}=\frac{2}{b}$

So $a,b,c$ are in HP.

Hence option C is the answer.