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