Question

If $x,y,z$ are real numbers satisfying the expression $25(9x^2+y^2)+9z^2-15(5xy+yz+3xz)=0$, then $x,y,z$ are in

• A. AP
• B. GP
• C. HP
• D. AGP

Answer 1

The correct option is A

AP

Explanation for the correct option.

Step 1: Simplify the given equation.

$\begin{array}{rcl}25(9x^2+y^2)+9z^2-15(5xy+yz+3xz)& =& 0\\ \Rightarrow 225x^2+25y^2+9z^2-75xy-15yz-45xz& =& 0\\ & \Rightarrow & {\left(15x\right)}^2+{\left(5y\right)}^2+{\left(3z\right)}^2-5y\left(15x\right)-5y\left(3z\right)-3z\left(15x\right)=0\\ \Rightarrow \frac{1}{2}\left[{\left(15x-5y\right)}^2+{\left(5y-3z\right)}^2+{\left(3z-15x\right)}^2\right]& =& 0\left[by{a}^2+b^2+c^2-ab-bc-ca=\frac{1}{2}\left[{\left(a-b\right)}^2+{\left(b-c\right)}^2+{\left(c-a\right)}^2\right]\right]\\ \Rightarrow {\left(15x-5y\right)}^2+{\left(5y-3z\right)}^2+{\left(3z-15x\right)}^2& =& 0\end{array}$

Step 2:Find the value of $x,y\mathrm{and}z$.

As, $x,y,z$ are real numbers, so

$$\begin{gathered} {\left(15x-5y\right)}^2=0 \\ \Rightarrow 15x=5y \\ \Rightarrow 3x=y \\ {\left(5y-3z\right)}^2=0 \\ \Rightarrow 5y=3z \\ \Rightarrow 5\left(3x\right)=3z \\ \Rightarrow z=5x \\ {\left(3z-15x\right)}^2=0 \\ \Rightarrow 3z=15x \\ \Rightarrow 3\left(5x\right)=15x \\ \Rightarrow x=1 \end{gathered}$$

So, $x=1.y=3,\mathrm{and}z=5$

Therefore, $x,y,z$ are in AP with $a=1,\mathrm{and}d=2$

Hence, option A is correct.