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 $\left(x^2-y^2+2 z-w\right)$.

Answer 1

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 $x =\frac{2 a }{ r }- a ; y =\frac{ a }{ r } ; z = a ; w = ar$
Now, $x+w=16 $ and $y+z=8$
$\Rightarrow \frac{2 a }{ r }- a + ar =16 $ and $ \frac{ a }{ r }+ a =8$
$\Rightarrow a\left(\frac{2}{r}-1+r\right)=16 $ and $ a(1+r)=8 r$
$\Rightarrow 8 r\left(\frac{2}{r}-1+r\right)=16(1+r) \Rightarrow 2-r+r^2=2+2 r$
$\therefore r ^2=3 r$
$r=0$ (reject) or 3
$\therefore a =6$
$\therefore x =-2 ; y =2 ; z =6 ; w =18$
Hence, $x^2-y^2+2 z-w=(-2)^2-(2)^2+2(6)-18=4-4+12-18=-6 \Rightarrow|-6|=6$