Question

If $x,y,z$ are real and $9x^2+16y^2+25z^2-12xy-20yz-15zx=0,$ then $\frac{1}{x},\frac{1}{y},\frac{1}{z}$ are in

• A. arithmetic progression
• B. geometric progression
• C. harmonic progression
• D. None of the above

Answer 1

The correct option is A arithmetic progression

Given : $9x^2+16y^2+25z^2-12xy-20yz-15zx=0$
$\Rightarrow (3x{)}^2+(4y{)}^2+(5z{)}^2-(3x\left)\right(4y)-(4y\left)\right(5z)-(5z\left)\right(3x)=0\cdots (i)$
It is of the form $a^2+b^2+c^2-ab-bc-ac=0$
which can be written as
$\frac{1}{2}\left[\right(a-b{)}^2+(b-c{)}^2+(c-a{)}^2]=0$
$\Rightarrow a=b=c$

Similarly, equation $\left(i\right)$ can be written as
$(3x-4y{)}^2+(4y-5z{)}^2+(5z-3x{)}^2=0$
$\Rightarrow 3x=4y=5z$
Let $3x=4y=5z=k$
$\Rightarrow x=\frac{k}{3},y=\frac{k}{4},z=\frac{k}{5}$
$\therefore \frac{1}{x},\frac{1}{y},\frac{1}{z}$ are in A.P.