Question

Let $x,y,z,w$ be four non-zero real numbers such that $x,y,z$ (in order) are in arithmetic progression and $y,z,w$ (in order) are in geometric progression If $x+w=16$ and $y+z=8$, then find the absolute value of $\left(x^2-y^2+2z-w\right)$.

Answer 1

Answer: 6

Given $x,y,z,w$ are 4 numbers such that $x,y,z$ are in A.P. and $y,z,w$ are in G.P.
Let $x=\frac{2a}{r}-a;y=\frac{a}{r};z=a;w=ar$
Now, $x+w=16$ and $y+z=8$
$\Rightarrow \frac{2a}{r}-a+ar=16$ and $\frac{a}{r}+a=8$
$\Rightarrow a\left(\frac{2}{r}-1+r\right)=16$ and $a(1+r)=8r$
$\Rightarrow 8r\left(\frac{2}{r}-1+r\right)=16(1+r)\Rightarrow 2-r+r^2=2+2r$
$\therefore r^2=3r$
$r=0$ (reject) or 3
$\therefore a=6$
$\therefore x=-2;y=2;z=6;w=18$
Hence, $x^2-y^2+2z-w=(-2{)}^2-(2{)}^2+2(6)-18=4-4+12-18=-6\Rightarrow |-6|=6$