Question

If $(y-x),2(y-a)\mathrm{and}(y-z)$ are in H.P., then $x-a,y-a,z-a$ are in:

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

Answer 1

The correct option is B

G.P

Explanation for the correct answer.

Finding the relation:

As, $(y-x),2(y-a)\mathrm{and}(y-z)$ are in H.P, so $\frac{1}{(y-x)},\frac{1}{2(y-a)}\mathrm{and}\frac{1}{(y-z)}$ are in A.P. So,

$\begin{array}{rcl}\frac{1}{2(y-a)}-\frac{1}{(y-x)}& =& \frac{1}{(y-z)}-\frac{1}{2(y-a)}\\ & \Rightarrow & \frac{1}{2(y-a)}+\frac{1}{2(y-a)}=\frac{1}{(y-z)}+\frac{1}{(y-x)}\\ \Rightarrow \frac{2}{2(y-a)}& =& \frac{(y-x)+(y-z)}{(y-x)(y-z)}\\ \Rightarrow \frac{1}{(y-a)}& =& \frac{2y-x-z}{(y-x)(y-z)}\\ & \Rightarrow & (y-x)(y-z)=(y-a)\left(2y-x-z\right)\\ \Rightarrow y^2-yz-xy+xz& =& 2y^2-xy-yz-2ay+ax+az\\ \Rightarrow xz-ax-az& =& y^2-2ay\end{array}$

By adding $a^2$ on both sides, we get

$\begin{array}{rcl}xz-ax-az+a^2& =& y^2-2ay+a^2\\ & \Rightarrow & z\left(x-a\right)-a\left(x-a\right)={\left(y-a\right)}^2\\ \Rightarrow \left(z-a\right)\left(x-a\right)& =& {\left(y-a\right)}^2\end{array}$

This shows that $x-a,y-a,z-a$ are in G.P

Hence, option B is correct.