Question

If $2(y-a)$ is the H.M. between $y-x$ and $y-z$ then $x-a,y-a,z-a$ are in?

• A. A.P.
• B. G.P.
• C. H.P.
• D. None of these

Answer 1

Answer: (B)

G.P.

We know that if $x$, y are in HP then $HM=S_{xy}$

$x+y$

$f(y-a)=\frac{x(y-x)(y-z)}{2y-(x+z)}$

$2y^2-2ay-x4y+ax-zy+az=y^2-yz-xy+xz$

$y^2-2ay+ax=z(x-a)$

add and subtract a from $L\cdot H\cdot S$

$y^2-2ay+a^2+ax-a^2=z(x-a)$

$(y-a{)}^2+a(x-a)=z(x-a)$

$(y-a{)}^2=(z-a\left)\right(x-a)$

$\Rightarrow x-a,y-a,z-a$ are in $GP$

Option $\left(b\right)$