Question

Three numbers $x, y$ and $z $ are in arithmetic progression. If $x + y + z = - 3$ and $xyz= 8$, then $x^2 + y^2 + z^2$ is equal to

• A. 9
• B. 10
• C. 21
• D. 20
• E. 1

Answer 1

Answer: (C)

Let $x=a-r, y=a, z=a +r$
Now, we have
$x+y+z=-3$
$\therefore a- r+ a+ a +r=-3$
$\Rightarrow 3 a=-3$
$\Rightarrow a=-1$
Again, $x y z=8$
$\therefore (a-r)(a)(a +r)=8$
$\Rightarrow a\left(a^{2}-r^{2}\right)=8$
$\Rightarrow -1\left(1-r^{2}\right)=8$
$\Rightarrow -1+r^{2}=8 $
$\Rightarrow r^{2}=9$
$\Rightarrow r=\pm 3 $
$\therefore x, y, z$ are $-4,-1,2$ or $2,-1,-4 $
$\therefore x^{2}+y^{2}+z^{2}=(-4)^{2}+(-1)^{2}+(2)^{2} $
$ =16+1+4=21$